APR 23 !W5 



APPLIED MECHANICS. 



BY 



GAETANO LANZA, S.B., C. & M.E., 

PROFESSOR ° F THE ° R r; s C - *» ™> -„ AN ,CS, „ A SS AC „„SB TTS 
INSTITUTE OF TECHNOLOGY. 



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NEW YORK : 
JOHN WILEY & SONS 

1 S ASTOR PLACE. 
I88 5 . 



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Copyright, 1885, 
By GAETANO LANZA. 



ELECTROTYPED AND PRINTED 

BY RAND, AVERY, AND COMPANY, 

BOSTON. 



V » 



PREFACE. 



This book is the result of the experience of the writer 
in teaching the subject of Applied Mechanics for the last 
twelve years at the Massachusetts Institute of Technology. 

The immediate object of publishing it is, to enable him to 
dispense with giving to the students a large amount of notes. 
As, however, it is believed that it may be found useful by 
others, the following remarks in regard to its general plan 
are submitted. 

The work is essentially a treatise on strength and stabil- 
ity ; but, inasmuch as it contains some other matter, it was 
thought best to call it " Applied Mechanics," notwithstanding 
the fact that a number of subjects usually included in trea- 
tises on applied mechanics are omitted. 

It is primarily a text-book ; and hence the writer has endeav- 
ored to present the different subjects in such a way as 
seemed to him best for the progress of the class, even though 
it be at some sacrifice of a logical order of topics. While 
no attempt has been made at originality, it is believed that 
some features of the work are quite different from all pre- 



IV PREFACE. 

vious efforts ; and a few of these cases will be referred to, 
with the reasons for so treating them. 

In the discussion upon the definition of "force," the object 
is, to make plain to the student the modern objections to the 
usual ways of treating the subject, so that he may have a 
clear conception of the modern aspect of the question, rather 
than to support the author's definition, as he is fully aware 
that this, as well as all others that have been given, is open 
to objection. 

In connection with the treatment of statical couples, it 
was thought best to present to the student the actual effect 
of the action of forces on a rigid body, and not to delay this 
subject until dynamics of rigid bodies is treated, as is usually 
done. 

In the common theory of beams, the author has tried to 
make plain the assumptions on which it is based. A little 
more prominence than usual has also been given to the longi- 
tudinal shearing of beams. 

In that part of the book that relates to the experimental 
results on strength and elasticity, the writer has endeavored 
to give the most reliable results, and to emphasize the fact, 
that, to obtain constants suitable for use in practice, we 
must deduce them from tests on full-size pieces. This prin- 
ciple of being careful not to apply experimental results to 
cases very different from those experimented upon, has long 
been recognized in physics, and therefore needs no justifica- 
tion. 

The government reports of tests made at the Watertown 
Arsenal have been extensively quoted from, as it is believed 



PREFACE. 



that they furnish some of our most reliable information on 
these subjects. 

The treatment of the strength of timber will be found to 
be quite different from what is usually given ; but it speaks 
for itself, and will not be commented upon here. 

In the chapter on the " Theory of Elasticity," a combina- 
tion is made of the methods of Rankine and of Grashof. 

In preparing the work, the author has naturally consulted 
the greater part of the usual literature on these subjects ; and, 
whenever he has drawn from other books, he has endeavored to 
acknowledge it. He wishes here to acknowledge the assist- 
ance furnished him by Professor C. H. Peabody of the Massa- 
chusetts Institute of Technology, who has read all the proofs, 
and has aided him materially in other ways in getting out the 
work. 

Gaetano Lanza. 

Massachusetts Institute of Technology, 
April, 1885. 



TABLE OF CONTENTS. 



CHAPTER I. 
Composition and Resolution of Forces i 

CHAPTER II. 
Dynamics 75 

CHAPTER III. 
Roof-Trusses 138 

CHAPTER IV. 
Bridge-Trusses 174 

CHAPTER V. 
Centre of Gravity 211 

CHAPTER VI. 
Strength of Materials 230 

CHAPTER VII. 
Strength of Materials as Determined by Experiment 329 

vii 



Vlll TABLE OF CONTENTS. 

CHAPTER VIII. 
Continuous Girders , . . 555 

CHAPTER IX. 
Equilibrium Curves. — Arches and Domes 591 

CHAPTER X. 
Theory of Elasticity, and Applications 658 



APPLIED MECHANICS. 



CHAPTER I. 

COMPOSITION AND RESOLUTION OF FORCES. 

§ i. Fundamental Conceptions. — The fundamental con- 
ceptions of Mechanics are Force, Matter, Space, Time, and 
Motion. 

§2. Relativity of Motion. — The limitations of our natures 
are such that all our quantitative conceptions are relative. 
The truth of this statement may be illustrated, in the case of 
motion, by the fact, that, if we assume the shore as fixed in 
position, a ship sailing on the ocean is in motion, and a ship 
moored in the dock is at rest ; whereas, if we assume the sun 
as our fixed point, both ships are really in motion, as both par- 
take of the motion of the earth. We have, moreover, no means 
of determining whether any given point is absolutely fixed in 
position, nor whether any given direction is an absolutely fixed 
direction. Our only way of determining direction is by means 
of two points assumed as fixed ; and the straight line joining 
them, we are accustomed to assume as fixed in direction. 
Thus, it is very customary to assume the straight line joining 
the sun with any fixed star as a line fixed in direction ; but if 
the whole visible universe were in motion, so as to change the 
absolute direction of this line, we should have no means of 
recognizing it. 



APPLIED MECHANICS. 



§3. Rest and Motion. — In order to define rest and 
motion, we have the following; viz., — 

When a single point is spoken of as having motion or rest, 
some other point is always expressed or understood, which is 
for the time being considered as a fixed point, and some direc- 
tion is assumed as a fixed direction : and we then say that the 
first-named point is at rest relatively to the fixed point, when 
the straight line joining it with the fixed point changes neither 
in length, nor in direction; whereas it is said to be in motion 
relatively to the fixed point, when this straight line changes in 
length, in direction, or in both. 

If, on the other hand, we had considered the first-named 
point as our fixed point, the same conditions would determine 
whether the second was at rest, or in motion, relatively to the 
first. 

A body is said to be at rest relatively to a given point and 
to a given direction, when all its points are at rest relatively to 
this point and this direction. 

§4. Velocity. — When the motion of one point relatively 
to another, or of one body relatively to another, is such that it 
describes equal distances in equal times, however small be the 
parts into which the time is divided, the motion is said to be 
uniform and the velocity constant. 

The velocity, in this case, is the space passed over in a unit 
of time, and is to be found by dividing the space passed over in 
any given time by the time ; thus, if s represent the space 
passed over in time /, and v represent the velocity, we shall 

have 

s 

v = -. 
/ 

When the motion is not uniform, if we divide the time into 
small parts, and then divide the space passed over in one of 
these intervals by the time, and then pass to the limit as these 
intervals of time become shorter, we shall obtain the velocity. 



FORCE. 3 

Thus, if As represent the space passed over in the interval of 
time At, then we shall have 

V = limit of — as At diminishes, 
At 

or 

ds 

v = — . 
dt 

§ 5. Force. — We shall next attempt to obtain a correct defi- 
nition of force, or at least of what is called force in mechanics. 

It may seem strange that it should be necessary to do this ; 
as it would appear that clear and correct definitions must have 
been necessary in order to make correct deductions, and there- 
fore that there ought to be no dispute whatever over the mean- 
ing of the word force. Nevertheless, it is a fact in mechanics, 
as well as in all those sciences which attempt to deal with the 
facts and laws of nature, that correct definitions are only gradu- 
ally developed, and that, starting with very imperfect and often 
erroneous views of natural laws and phenomena, it is only after 
these errors have been ascertained and corrected by a long 
range of observation and experiment, and an increased range of 
knowledge has been acquired, that exactness and perspicuity 
can be obtained in the definitions. 

Now, this is precisely what has happened in the case of 
force. 

In ancient times rest was supposed to be the natural state 
of bodies ; and it was assumed that, in order to make them 
move, force was necessary, and that even after they had been 
set in motion their own innate inertia or sluggishness would 
cause them to come to rest unless they were constantly urged 
on by the application of some force, the bodies coming to rest 
whenever the force ceased acting. 

It was under the influence of these vague notions that such 
terms arose as Force of Inertia, Moment of Inertia, Vis Viva 
or Living Force, etc. 



AT PLIED MECHANICS. 



A number of these terms are still used in mechanics ; but in 
all such cases they have been re-defined, such new meanings 
having been attached to them as will bring them into accord 
with the more advanced ideas of the present time. Such defini- 
tions will be given in the course of this work, as the necessity 
may arise for the use of the terms. 

Moreover, it is to be regretted that there still prevails, to 
some extent, in the more popular class of scientific literature, a 
loose usage of such terms, which is very liable to impart erro- 
neous ideas to those whose minds are not clear as to their true 
meanings. 

NEWTON'S FIRST LAW OF MOTION. 

Ideas becoming more precise, in course of time there was 
framed Newton's first law of motion ; and this law is as fol- 
lows : — 

A body at rest will remain at rest, and a body in motion will 
continue to move uniformly and in a straight line, unless and 
until some external force acts upon it. 

The assumed truth of this law was based upon the observed 
facts of nature ; viz., — 

When bodies were seen to be at rest, and from rest passed 
into a state of motion, it was always possible to assign some 
cause ; i.e., they had been brought into some new relationship, 
either with the earth, or with some other body: and to this 
cause could be assigned the change of state from rest to motion. 
On the other hand, in the case of bodies in motion, it was seen, 
that, if a body altered its motion from a uniform rectilinear 
motion, there was always some such cause that could be 
assigned. Thus, in the case of a ball thrown from the hand, 
the attraction of the earth and the resistance of the air soon 
caused it to come to rest. In the case of a ball rolled along 
the ground, friction (i.e., the continual contact and collision with 
the ground) gradually destroyed its motion, and brought it to 



FORCE. 5 

rest ; whereas, when such resistances were diminished by rolling 
it on glass or on the ice, the motion always continued longer : 
hence it was inferred, that, were these resistances entirely 
removed, the motion would continue forever. 

In accordance with these views, the definition of force 
usually given was substantially as follows : — 

Force is that which causes, or tejids to cause, a body to change 
its state from rest to motion, from motion to rest, or to change its 
motion as to direction or speed. 

Under these views, uniform rectilinear motion was recog- 
nized as being just as much a condition of equilibrium, or of 
the action of no force or of balanced forces, as rest ; and the 
recognition of this one fact upset many false notions, destroyed 
many incorrect conclusions, and first rendered possible a science 
of mechanics. Along with the above-stated definition of force 
is ordinarily given the following proposition ; viz., — 

Forces are proportional to the velocities that they impart, in a 
unit of time, to the same body. The reasoning given in support 
of this proposition is as follows : — 

Suppose a body to be moving uniformly and in a straight 
line, and suppose a force to act upon it for a certain length of 
time t in the direction of the body's motion : the effect of the 
force is to alter the velocity of the body ; and it is only by this 
alteration of velocity that we recognize the action of the force. 
Hence, as long as the alteration continues at the same rate, we 
recognize the same force as acting. 

If, therefore, / represent the amount of velocity which the 
force would impart in one unit of time, the total increase in 
the velocity of the body will be ft; and, if the force now stop 
acting, the body will again move uniformly and in the same 
direction, but with a velocity greater by//. 

Hence, if we are to measure forces by their effects, it will 
follow that — 

The velocity which a force will impart to a given {or standard) 



6 APPLIED MECHANICS. 

body in a unit of time is a proper measure of the force. And 
we shall have, that two forces, each of which will impart the 
same velocity to the same body in a unit of time, are equal to 
each other ; and a force which will impart to a given body twice 
the velocity per unit of time that another force will impart to 
the same body, is itself twice as great, or, in other words, — 

Forces are proportio7ial to the velocities that they impart ', in a 
unit of time, to the same body. 

MODERN CRITICISM OF THE ABOVE. 

The scientists and the metaphysicians of the present time 
are recognizing two other facts not hitherto recognized, and the 
result is a criticism adverse to the above-stated definition of 
force. . Other definitions have, in consequence, been proposed ; 
but none are free from objection on logical grounds, and at the 
same time capable of use in mechanics in a quantitative way. 

The two facts referred to are the following ; viz., — 

i°. That all our ideas of space, time, rest, motion, and even 
of direction, are, relative. 

2°. That, because two effects ar£ identical, it does not follow 
that the causes producing those effects are identical. 

Hence, in the light of these two facts, it is plain, that, inas- 
much as we can only recognize motion as relative, we can only 
recognize force as acting when at least two bodies are con- 
cerned in the transaction ; and also that if the forces are simply 
the causes of the motion in the ordinary popular sense of the 
word cause, we cannot assume, that, when the effects are equal, 
the causes are in every way identical, although we have, of 
course, a perfect right to say that they are identical so far as 
the production of motion is concerned. 

I shall now proceed, in the light of the above, to deduce a 
definition of force, which, although not free from objection, 
seems as free as any that has been framed. 

It is one of the facts of nature, that, when bodies are by any 



FORCE. 



means brought under certain relations to each other, certain 
tendencies are developed, which, if not interfered with, will 
exhibit themselves in the occurrence of certain definite phe- 
nomena. What these phenomena are, depends upon the nature 
of the bodies concerned, and on the relationships into which 
they are brought. 

As an illustration, we know that if an apple is placed at a 
certain height above the surface of the earth, there is developed 
between the two bodies a tendency to approach each other; 
and if there is no interference with this tendency, it exhihits 
itself in the fall of the apple. If, on the other hand, the apple 
were hung on the hook of a spring balance in the same posi- 
tion as before, the spring would stretch, and there would be 
developed a tendency of the spring to make the apple move 
upwards. This tendency to make the apple move upwards 
would be just equal to the tendency of the earth and apple to 
approach each other. This would be expressed by saying that 
the pull of the spring is just equal and opposite to the weight 
of the apple. 

As other illustrations of these tendencies developed in 
bodies when placed in certain relations to each other, we have 
the following cases : — 

(a) When two bodies collide. 

(b) When two substances, coming together, form a chemical 
union, as sodium and water. 

(c) When the chemical union is entered into only by raising 
the temperature to some special point. 

Any of these tendencies that are developed by bringing 
about any of these special relationships between bodies might 
properly be called a force ; and the term might properly be, and 
is, used in the same sense in the mental and moral world, as 
well as in the physical. In mechanics, however, we have to 
deal only with the relative motion of bodies ; and hence we 
give the name force only to tendencies to change the relative 



8 APPLIED MECHANICS. 

motion of the bodies concerned ; and this, whether these ten- 
dencies are unresisted, and exhibit themselves in the actual 
occurrence of a change of motion, or whether they are resisted 
by equal and opposite tendencies, and exhibit themselves in 
the production of a tensile, compressive, or other stress in the 
bodies concerned, instead of motion. 



DEFINITION OF FORCE. 

Hence our definition of force, as far as mechanics has to 
deal with it oris capable of dealing with it, is as follows; 
viz., — 

Force is a tendency to change the relative motion of the two 
bodies between which that tendency exists. 

Indeed, when, as in the illustration given a short time ago, 
the apple.is hung on the hook of a spring balance, there still 
exists a tendency of the apple and the earth to approach each 
other ; i.e., they are in the act of trying to approach each other; 
and it is this tendency, or act of trying ; that we call the force of 
gravitation. In the case cited, this tendency is balanced by 
an opposite tendency on the part of the spring; but, were the 
spring not there, the force of gravitation would cause the apple 
to fall. 

Professor Rankine calls force "an action between two bodies, 
either causing or tending to cause change in their relative rest 
or motion ;" and if the act of trying can be called an action, my 
definition is equivalent to his. 

For the benefit of any one who wishes to follow out the 
discussions that have lately taken place, I will enumerate the 
following articles that have been written on the subject : — 

(a) " Recent Advances in Physical Science," by P. G. Tait, 
Lecture XIV. 

(b) Herbert Spencer, " First Principles of Philosophy " 
(certain portions of the book). 



MEASURE OF FORCE. 



(c) Discussion by Messrs. Spencer and Tait, " Nature," Jan. 
2, 9, 1 6, 1879. 

(d) Force and Energy, "Nature," Nov. 25, Dec. 2, 9, 16, 
1880. 

§6. External Force. — We thus see, that, in order that a 
force may be developed, there must be two bodies concerned 
in the transaction ; and we should speak of the force as that 
developed or existing between the two bodies. 

But we may confine our attention wholly to the motion or 
condition of one of these two bodies ; and we may refer its 
motion either to the other body as a fixed point, or to some 
body different from either ; and then, in speaking of the force, 
we should speak of it as the force acting on the body under 
consideration, and call it an external force. It is the tendency 
of the other body to change the motion of the body under con- 
sideration relatively to the point considered as fixed. 

§7. Relativity of Force. — In adopting the above-stated 
definition of force, we acknowledge our incapacity to deal with 
it as an absolute quantity ; for we have defined it as a tendency 
to change the relative motion of a pair of bodies. Hence it is 
only through relative motion that we recognize force; and hence 
force is relative, as well as motion. 

§8. Newton's First Law of Motion. — In the light of 
the above discussion, we might express Newton's first law of 
motion as follows : — 

A body at rest, or in uniform rectilinear motion relatively to 
a given point assumed as fixed, will continue at rest, or in uni- 
form motion in the same direction, unless and until some external 
force acts either on the body in question, or on the fixed point, 
or 011 the body which furnishes us our fixed direction. This law 
is really superfluous, as it has all been embodied in the defini- 
tion. 

§9. Measure of Force. — We next need some means of 
comparing forces with each other in magnitude; and, subse- 



IO APPLIED MECHANICS. 

quently, we need to select one force as our unit force, by means 
of which to estimate the magnitude of other forces. 

Let us suppose a body moving uniformly and in a straight 
line, relatively to some fixed point ; as long as this motion 
continues, we recognize no unbalanced force acting on it ; 
but, if the motion changes, there must be a tendency to change 
that motion, or, in other words, an unbalanced force is acting 
on the body from the instant when it begins to change its 
motion. 

Suppose a body to be moving uniformly, and a force to be 
applied to it, and to act for a length of time /, and to be so applied 
as not to change the direction of motion of the body, but to 
increase its velocity; the result will be, that the velocity will be 
increased by equal amounts in equal times, and if f represent 
the amount of velocity the force would impart in one unit of 
time, the total increase in velocity will be //. This results 
merely from the definition of a force ; for if the velocity pro- 
duced in one (a standard) body by a given force is twice as 
great as that produced by another given force, then is the ten- 
dency to produce velocity twice as great in the first case as in 
the second, or, in other words, the first force is twice as great 
as the second. Hence — 

Forces are proportional to the velocities which they will impart 
to a given {or standard) body in a unit of time. 

We may thus, by using one standard body, determine a 
set of equal forces, and also the proportion between different 
forces. 

§10. Measure of Mass. — After having determined, as 
shown, a set of equal (unit) forces, if we apply two of them 
to different bodies, and let them act for the same length of time 
on each, and find that the resulting velocities are unequal, these 
bodies are said to have unequal masses ; whereas, if the result- 
ing velocities are equal, they are said to have equal masses. 

Hence we have the following definitions : — 



RELATION BETWEEN FORCE AND MOMENTUM. II 

1°. Equal forces are those which, by acting for eqtial times 
on the same or standard body, impart to it equal velocities. 

2°. Equal masses are those masses to which equ,al foires 
will impart equal velocities in equal times. 

§ ii. Suppose two bodies of equal mass moving side by 
side with the same velocity, and uniformly, let us apply to 
one of them a force F in the direction of the body's motion : 
the effect of this force is to increase the velocity with which the 
body moves ; and if we wish, at the same time, to increase 
the velocity of the other, so that they will continue to move 
side by side, it will be necessary to apply an equal force to that 
also. 

We are thus employing a force 2F to impart to the two 
bodies the required increment of velocity. 

If we unite them into one, it still requires a force 2F to 
impart to -the one body resulting from their union the re- 
quired increment of velocity : hence, if we double the mass 
to which we wish to impart a certain velocity, we must double 
the force, or, in other words, employ a force which would 
impart to the first mass alone a velocity double that required. 
Hence — 

Forces are proportional to the masses to which they will impart 
the same velocity in the same time. 

§12. Momentum. — The product obtained by multiplying 
the number of units of mass in a body by its velocity is called 
the momentum of the body. 

§13. Relation between Force and Momentum. — The 
number of units of momentum imparted to a body in a unit of 
time by a given force, is evidently identical with the number 
of units of velocity that would be imparted by the same force, 
in the same time, to a unit mass. Hence — 

Forces are proportional to the momenta {or velocities per unit 
of mass) which they will gejterate in a unit of time. 



12 APPLIED MECHANICS. 

Hence, if F represent a force which generates, in a unit of 
time, a velocity f in a body whose mass is m, we shall have 

Ftxmf; 

and, inasmuch as the choice of our units is still under our con- 
trol, we so choose them that 

F= mf; 

i.e., the force F contains as many units of force as mf contains 
units of momentum ; in other words, — 

The momentum generated in a body in a unit of time by a 
force acting in the direction of the body s motion, is taken as 
a measure of the force. 

§14. Statical Measure of Force. — When the forces are 
prevented from producing motion by being resisted by equal 
and opposite forces, as is the case in that part of mechanics 
known as Statics, they must be measured by a direct comparison 
with other forces. An illustration of this has already been 
given in the case of an apple hung on the hook of a spring 
balance. In that case the pull of the spring is equal in magni- 
tude to the weight of the apple : indeed, it is very customary 
to adopt for forces what is known as the gravity measure, in 
which case we take as our unit the gravitation, or tendency to 
fall, of a given piece of metal, at a given place on the surface 
of the earth ; in other words, its weight at a given place. 

The gravity unit may thus be the kilogram, the pound, or 
the ounce, etc. 

It is evident, moreover, from our definition of force, and the 
subsequent discussion, that whatever we take as our unit of 
mass, the statical measure of a force is proportional to its 
dynamical measure; i.e., the numbers representing the magni- 
tudes of any two forces, in pounds, are proportional to the 
momenta they will impart to any body in a unit of time. 

§15. Gravity Measure of Mass. — If we assume one 
pound as our unit of force, one foot as our unit of length, and 



NEWTON'S SECOND LAW OF MOTION 1 3 

one second as our unit of time, the ratio between the number 
of pounds in any given force and the momentum it will impart 
to a body on which it acts unresisted for a unit of time, will 
depend on our unit of mass ; and, as we are still at liberty to fix 
this as we please, it will be most convenient so to choose it 
that the above-stated ratio shall be unity, so that there shall be 
no difference in the measure of a force, whether it is measured 
statically or dynamically. Now, it is known that a body falling 
freely under the action of its own weight acquires, every second, 
a velocity of about thirty-two feet per second : this number is 
denoted by g, and varies for different distances from the centre 
of the earth, as does also the weight of the body. 

Now, if W represent the weight of the body in pounds, and 
m the number of units of mass in its mass, we must have, in 
order that the statical and dynamical measures may be equal, 

W — mg. 
Hence 

W 

m = — ; 

g 

i.e., the number of units of mass in a body is obtained by divid- 
ing the weight in pounds, by the value of g at the place where 
the weight is determined. 

The values of W and of g vary for different positions, but 
the value of m remains always the same for the same body. 

UNIT OF MASS. 

If m = 1, then W — g; or, in words, — 

The weight in pounds of the unit of mass {when the gravity 
measure is used) is equal to the value of g in feet per second for 
the same place. 

§ 16. Newton's Second Law of Motion. — Newton's 
second law of motion is as follows : — 



APPLIED MECHANICS. 



" Change of momentum is proportional to the impressed mov- 
ing force, and occurs along the straigJit line in which the force is 
impressed'' 

Newton states further in his " Principia :" — 

"If any force generate any momentum, a double force 
will generate a double, a triple force will generate a triple, 
momentum, whether simultaneously and suddenly, or gradually 
and successively impressed. And if the body was moving 
before, this momentum, if in the same direction as the motion, 
is added ; if opposite, is subtracted ; or if in an oblique direc- 
tion, is annexed obliquely, and compounded with it, according 
to the direction and magnitude of the two." 

Part of this law has reference to the proportionality between 
the force and the momentum imparted to the body ; and this 
has been already embodied in our definition of force, and illus- 
trated in the discussion on the measure of forces. 

The other part is properly a law of motion, and may be 
expressed as follows : — 

If a body have tzvo or more velocities imparted to it simulta- 
neously, it zvill move so as to preserve them all. 

The proof of this law depends merely upon a proper con- 
ception of motion. To illustrate this law when two velocities 
are imparted simultaneously to a body, let us suppose a man 
walking on the deck of a moving ship : he then has two motions 
in relation to the shore, his own and that of the ship. 

Suppose him to walk in the direction of motion of the 
ship at the rate of 10 feet per second, while the ship moves at 
25 feet per second relatively to the shore : then his motion in 
relation to the shore will be 25 + 10 = 35 feet per second. 
If, on the other hand, he is walking in the opposite direction at 
the same rate, his motion relatively to the shore will be 25 — 
10 = 15 feet per second. 

Suppose a body situated at A (Fig. 1) to have two motions 
imparted to it simultaneously, one of which would carry it to B 




POLYGON OF MOTIONS 1 5 

in one second, and the other to C in one second ; and that it is 
required to find where it will be at the end of one second, and 
what path it will have pursued. 

Imagine the body to move in obedience 
to the first alone, during one second : it 
would thus arrive at B ; then suppose the 
second motion to be/ imparted to the body, 
instead of the first, it will arrive at the end of the next sec- 
ond at D, where BD is equal and parallel to AC. When 
the two motions are imparted simultaneously, instead of suc- 
cessively, the same point D will be reached in one second, 
instead of two; and by dividing AB and AC into the same 
(any) number of equal parts, we can prove that the body will 
always be situated at some point of the diagonal AD of the 
parallelogram, hence that it moves along AD. Hence follows 
the proposition known as the parallelogram of motions. 

PARALLELOGRAM OF MOTIONS. 

If there be simultaneously impressed on a body two velocities, 
which would separately be represented by the lines AB and AC, 
the actual velocity will be represented by the line AD, which is 
the diagonal of the parallelogram of which AB and AC are the 
adjacent sides. 

§17. Polygon of Motions. — In all the above cases, the 
point reached by the body at the end of a second when the 
two motions take place simultaneously is the same as that which 
would be reached at the end of two seconds if the motions took 
place successively ; and the path described is the straight line 
joining the initial position of the body, with its position at the 
end of one second when the motions are simultaneous. 

The same principle applies whatever be the number of 
velocities that may be imparted to a body simultaneously. 
Thus, if we suppose the several velocities imparted to be 
(Fig. 2) AB, AC, AD, AE, and AF, and it be required to 



1 6 APPLIED MECHANICS. 



determine the resultant velocity, we first let the body move 

A with the velocity AB for one second ; at the 

~~^-^P end of that second it is found at B ; then let 

* c Nv it move with the velocity AC only, and at 

/ the end of another second it will be found 

/ at c ; then with AD only, and at the end of 

/ the third second it will be found at d; at the 

/ end of the fourth at e ; at the end of the fifth 

^e at /. Hence the resultant velocity, when all 

FlG - 2 - are imparted simultaneously, is Af or the 

closing side of the polygon. 

This proposition is known as the polygon of motions. 

POLYGON OF MOTIONS. 

If there be simultaneously impressed on a body any number 
of velocities, the resulting velocity will be represented by the 
closing side of a polygon of which the lines representing tlie 
separate velocities form the other sides. 

§ 1 8. Characteristics of a Force. — A force has three 
characteristics, which, when known, determine it ; viz., Point 
of Application, Direction, and Magnitude. These can be repre- 
sented by a straight line, whose length is made proportional to 
the magnitude of the force, whose direction is that of the 
motion which the force imparts, or tends to impart, and one end 
of which is the point of application of the force ; an arrow-head 
being usually employed to indicate the direction in which the 
force acts. 

§ 19. Parallelogram of Forces. 

Proposition. — If two forces acting simultaneo7isly at the 
same point be represented, in point of application, dii'ection, 
and magnitude, by two adjacent sides of a parallelogram, their 
resultant will be represented by the diagonal of the parallelo- 
gram, dratvn from the point of application of the two forces. 

Proof. — In the last part of § 16 was proved the propo- 



PARALLELOGRAM OF FORCES. \J 

sition known as the Parallelogram of Motions, for the state- 
ment of which the reader is referred to the close of that 
section. 

We have also seen that forces are proportional to the velo- 
cities which they impart, or tend to impart, in a unit of time, 
to the same body. 

Hence the lines representing the two impressed forces are 
coincident in direction with, and proportional to, the lines repre- 
senting the velocities they would impart in a unit of time to 
the same body ; and moreover, since the resultant velocity is 
represented by the diagonal of the parallelogram drawn with 
the component velocities as sides, the resultant force must coin- 
cide in direction with the resultant velocity, and the length of 
the line representing the resultant force will bear to the result 
ant velocity the same ratio that one of the component forces 
bears to the corresponding velocity. Hence it follows, that the 
resultant force will be represented by the diagonal of the paral- 
lelogram having for sides the two component forces. 

§ 20. Parallelogram of Forces : Algebraic Solution. 

Problem. — Given two forces F and F x acting at tlie same 
point A {Fig. 3), and inclined to each other at an angle 6; required 
the magnitude and direction of the resultant 
force. 

Let AC represent F, AB represent F s , 
and let angle BAC — 0; then will R — AD 
represent in magnitude and direction the 
resultant force. Also let angle DAC = a; then from the tri- 
angle DAC we have 

AD = AC 2 + CD* - 2AC. CDcosACD. 




But 



ACD = 180 - 6 .'. cosACD=-cos$ 
R* = F* + F* + 2FF X cos 



R = s/F* + F* + *FF X cos $. 



i8 



APPLIED MECHANICS. 



This determines the magnitude of R. To determine its direc- 
tion, let angle CAD = a .*. angle BAD = — a, and we 
shall have from the triangle DAC 



or 



and similarly 



CD : AD = sin CAD : sin ACD, 
F t : R == sin a : sin 6 



sin a = — sin 0, 
R 



■p 

sin(# — a) = — sin0. 
v } R 



EXAMPLES. 

i°. Given F = 4 7-34, ^. = 7546, = 73 ° 14' 21" 
2 . Given ^ = 5.36, F t = 4-27, = 3 2 ° IO ' 

3 . Given F = 42.00, 7% = 31.00, = 150 

4 . Given F = 47-°°; ^1 = 75-°°> # = 2 53° 



find 7? and a. 
find j? and a. 
find R and a. 
find R and a. 



§21. Parallelogram of Forces when 0= 90 . — When 
the two given forces are at right angles to each other, the for- 
mulae become very much simplified, since the parallelogram 
becomes a rectangle. 

From Fig. 4 we at once deduce 




R = 


sIf* 


+ ^ 2 , 


sin a = 


F T 
R' 




COS a = 


F 
R 




EXAMPLES. 







Given F — 3.0, F l = 5.0 ; find R and a. 
Given F = 3.0, F x — —5.0 ; find A* and a. 
Given F — 5.0, F x = 12.0 ; find ./? and a. 
Given F — 23.2, F t = 21.3 ; find ^ and a. 



DECOMPOSITION OF FORCES IN ONE PLANE. 1 9 

§ 22. Triangle of Forces. — If three forces be represented, 
in magnitude and direction, by the three sides of a triangle taken 
in order, tlien, if these forces be simultaneously applied at 07ie 
point, they will balance each other. 

Convej'sely, three forces zvhich, when simultaneously applied 
at one point, balance each otlier, can be correctly represented in 
magnitude and direction by the three sides of a triangle taken in 
order. 

These propositions, which find a very extensive application, 
especially in the determination of the stresses in roof and 
bridge trusses, are proved as follows : — 

If we have two forces, AC and AB (see Fig. 3), acting at the 
point A, their resultant is, as we have already seen, AD ; and 
hence a force equal in magnitude and opposite in direction to 
AD will balance the two forces AC and AB. Now, the sides of 
the triangle ACDA, if taken in order, represent in magnitude 
and direction the force AC, the force CD or AB, and a force 
equal and opposite to AD; and these three forces, if applied at 
the same point, would balance each other. Hence follows the 
proposition. 

Moreover, we have 

AC : CD : DA = sin ADC : sin CAD : smACD, 
or 

F : B, : R = sin (0 — a) : sin a : sin 6 ; 

or each force is, in this case, proportional to the sine of the 
angle between the other two. 

§ 23. Decomposition of Forces in one Plane. — It is 
often convenient to resolve a force into two components, in two 
given directions in a plane containing the force. Thus, suppose 
we have the force R = AD (Fig. 3), and we wish to resolve it 
into two components acting respectively in the directions AC 
and AB ; i.e., we wish to find two forces acting respectively in 
these directions, of which AD shall be the resultant : we 



20 



APPLIED MECHANICS. 



determine these components graphically by drawing a parallelo- 
gram, of which AD shall be the diagonal, and whose sides shall 
have the directions AC and AB respectively. The algebraic 
values of the magnitudes of the compo- 
nents can be determined by solving the 
triangle ADC. In the case when the 
directions of the components are at right 
angles to each other, let the force R 
(Fig. 5), applied at O, make an angle a 
with OX. We may, by drawing the rect- 
angle shown in the figure, decompose R 
into two components, /^and F lt along OX and O Y respectively ; 
and we shall readily obtain from the figure, 



Fig. 5. 



F — R cos a, F Y = R sin a. 




Fig. 6. 



EXAMPLES. 

i°. The force exerted by the steam upon the piston of a steam-engine 

at the moment when ifis in the position shown in the figure is AB = 

1000 lbs. The resistance of the 

guides upon the cross-head DE is 

vertical. Determine the force acting 

along the connecting-rod AC and 

the pressure on the guides ; also 

resolve the force acting along the connecting-rod into 
two components, one along, and the 
other at right angles to, the crank OC. 
2 . A load of 500 lbs. is placed at 
the apex C of the frame ACB : find" 

the stresses in A C and CB respectively. 

3 . A load of 4000 lbs. is hung at C, on the crane 

ABC: find the pressure in the boom BC, and the pull 

on the tie AC, where BC makes an angle of 60° with the horizontal, 

and AC an angle of 15 . 





COMPOSITION OF FORCES IN ONE PLANE. 21 

4°. A force whose magnitude is 7 is resolved into two forces whose 
magnitudes are 5 and 3 : find the angles they make with the given 
force. 

§ 24. Composition of any Number of Forces in One 
Plane, all applied at the Same Point. 

(a) Graphical Solution. — Let the forces be represented 
(Fig. 2) by AB, AC, AD, AE, and AF respectively. Draw Be 
|| and = AC, cd || and == AD, de || and — AE, and ef || and == 
AF; then will Af represent the resultant of the five forces. 
This solution is to be deduced from 
§ 17 in the same way as § 19 is deduced 
from § 16. 

(b) Algebraic Solution. - — Let 
the given forces (Fig. 9), of which 
three are represented in the figure, be 
F, F lf F 2 , F v F 4 , etc. ; and let the angles 
made by these forces with the axis OX 
be a, a r , a 2 , a 3 , a 4 , etc., respectively. 
Resolve each of these forces into two components, in the 
directions OX and OY respectively. We shall obtain for the 
components along OX 

OA — i^cosa, OB — /'jcoso,, OC = F 2 cosa 2 , etc.; 

and for those along OY 

OAt = i^sina, OB x — i^sinaj, OC x = i^ 2 sina 2 , etc. 

These forces are equivalent to the following two ; viz., a 
force .Fcos a + F t cos a t -+■ F 2 cos a 2 -f- F 3 cos a 3 + etc. along OX, 
and a force j^sin a + F z sin a x -f- F 2 sin a 2 + F 3 sin a 3 + etc. along 
O Y. The first may be represented by XFcos a, and the second 
by ^Fsina, where 2 stands for algebraic stun. There remains 
only to find the resultant of these two, the magnitude of which 
is given by the equation 

R = V^XFcosa)* + (XFsina) 2 ; 




22 



APPLIED^ MECHANICS. 



and, if we denote by a r the angle made by the resultant with 
OX, we shall have 

X^sina 



S/^COSa 
cosa r = , sinay 

R 



R 



EXAMPLES. 



1^ = 47 

i°. Given \ F ' = 73 

^2 = 43 

,^3= 2 3 



a = 2 
a, = 48 



a, = 52 



a, = 112 



~ J Find the result- 
ant force and 
its direction. 



Solution. 



P. 


a. 


COS a. 


sin a. 


F COS a. 


F 'sin a. 


47 
73 
43 
23 


21° 

4 8° 

82° 

112° 


0.93358 

O.669I3 

O.I39I7 

-O.37461 


0.35 8 37 

0.743I5 
O.99O27 

O.92718 


43.87826 
48.84649 

5.9843 1 
-8.61603 


16.84339 

54.24995 
42.58161 
21.32414 










9O.O93O3 


134.99909 



.•. XFcosa = 90.09303, X^sina == 134.99909, 

.-. R = ^(^Fcosa) 2 4- (XFsina)* = 162.2976. 

log X/ 7 COS a. = 1 .95469 1. 

logi? = 2.210331 

log COS Or = 9.744360 

(v = 5 6 ° 17'- 

Observation. — It would be perfectly correct to use the minus sign 
in extracting the square root, or to call R = —162.2976 ; but then we 
should have 



cosu r 



_ 90.09303 



or 



— 162.2976' 
180 + 56 17' 



and 



sinar = _L3M22£2_, 

— 162.2976 

- 2 3 6°i 7 '; 



COMPOSITION OF FORCES APPLIED AT SAME POINT. 2$ 

a result which, if plotted, would give the same force as when we call 
R ■= 162.2976 and a r — 56 17'. 

Hence, since it is immaterial whether we use the plus or the minus sign 
in extracting the square root provided the ?-est of the computation be 
consistent with it, we shall, for co?ivenience, use always plus. 



2°. 


^=4, 


a = 77°, 




^= 3, 


a T = 82°, 




jF 2 = 10, 


a 2 — 1 63°, 




^3= 5> 


a 3 = 2 75°. 


3°. 


^=5, 


a = cos -I |-, 




F*= 4, 


«i = 0, 




f 2 = 3, 


a 2 = 90°. 



§25. Polygon of Forces. — If any number of forces be 
represented in magnitude and direction by the sides of a polygon 
taken in order, then, if these forces be simultaneously applied at 
one point, they will balance each other. 

Conversely, any number of forces which, when simultaneously 
applied at oiie point, balance eacli other, can be correctly repre- 
sented in magnitude and direction by the sides of a polygon taken 
in order. 

These propositions are to be deduced from § 24 (a) in the 
same way as the triangle of forces is deduced from the parallelo- 
gram of forces. 

§ 26. Composition of Forces all applied at the Same 
Point, and not confined to One Plane. — This problem can 
be solved by the polygon of forces, since there is nothing in 
the demonstration of that proposition that limits us to a plane 
rather than to a gauche polygon. 

The following method, however, enables us to determine 
algebraic values for the magnitude of the resultant and for its 
direction. 



24 



APPLIED MECHANICS. 




Fig. io. 



We first assume a system of three rectangular axes, OX, 

OY, and OZ (Fig. io), whose origin 
is at the common point of the given 
forces. Now, let OE = F be one 
of the given forces. First resolve 
it into two forces, OC and OD, the 
first of which lies in the z axis, and 
the second perpendicular to OZ, 
or, as it is usually called, in the z 
plane ; the plane perpendicular to 
OX being the x plane, and that 
perpendicular to OY the y plane. 
Then resolve OD into two com- 
ponents, OA along OX, and OB along OY. We thus obtain 
three forces, OA, OB, and OC respectively, which are equivalent 
to the single force OE. These three components are the edges 
of a rectangular parallelopiped, of which OE — Bis the diagonal. 
Let, now, 

angle BOX = a, BOY = /?, and EOZ = y ; 

and we have, from the right-angled triangles EOA, BOB, and 
EOC respectively, 

OA = B cos a, OB^Bcos/3, OC = B cosy. 
Moreover, 

OA 2 + OB 2 = OD 2 and OD 2 + OC 2 = OB 2 
.-. OA 2 + OB 2 + OC 2 = OB 2 , 

and by substituting the values of OA, OB, and OC, given above, 

we obtain 

cos 2 a -f cos 2 /3 -j- cos 2 y = i ; 



a purely geometrical relation existing between the three angles 
that any line makes with three rectangular co-ordinate axes. 

When two of the angles a, /?, and y are given, the third can 
be determined from the above equation. 



COMPOSITION OF FORCES APPLIED AT SAME POINT. 2$ 



Resolve, in the same way, each of the given forces into 
three components, along OX, OY, and OZ respectively, and we 
shall thus reduce our entire system 
of forces to the following three 
forces : — 

i°. A single force %F cos a along OX. 
2°. A single force ^Fcos/3 along OY. 
3°. A single force ^Fcosy along OZ. 

We next proceed to find a sin- 
gle resultant for these three forces. 



Let (Fig. ii) 



OA = 2F cos a 

OB = XFcos/?, 
OC = ZFcosy. 




Fig. ii. 



Compounding OA and OB, we find OD to be their resultant ; 
and this, compounded with OC, gives OE as the resultant of 
the entire system. Moreover, 



or 



OE* = OD* 4- OC 2 = OA* + 6>^ 2 + 6>C 2 , 
.tfz = (XFcostt) 2 -f (XFcos/?) 2 -f (XFcosy) 2 
.'. ^? = \/(XFcosa) 2 -|- (XFcos£) 2 + (XFcosy) 2 ; 



and if we let ii(?X = a „ EO Y — fi ri and EOZ = y r , we shall 
have 

OA %Fcosa 2Fcos(3 , %F cosy 

cos a r = = , cosB r — -, and cos y r = L . 

OE R ' H R ' r R 

This gives us the magnitude and direction of the resultant. 

The same observation applies to the sign of the radical for 
R as in the case of forces confined to one plane. 



26 



APPLIED MECHANICS. 



DETERMINATION OF THE THIRD ANGLE FOR ANY ONE FORCE. 

When two of the angles a, (3, and y are given, the cosine of 
the third may be determined from the equation, — 

COS 2 a + COS 2 /? + COS 2 y = I ; 

but, as we may use either the plus or the minus sign in extract- 
ing the square root, we have no means of knowing which of 
the two supplementary angles whose cosine has been deduced 
is to be used. 

Thus, suppose a = 45 , /3 = 6o°, then 



cosy 

•*• 7 



±Vi 
6o c 



2 
or i20 c 



±* 



but which of the two to use we have no means of deciding. 

This indetermination will be more clearly seen from the fol- 
lowing geometrical considerations : — 

The angle a (Fig. 12), being given as 45 , locates the line 

representing the force on a right 
circular cone, whose axis is OX, 
and whose semi-vertical angle is 
A OX = BOX — 45 . On the other 
hand, the statement that fi = 6o° 
locates the force on another right 
circular cone, having O Y for axis, 
and a semi-vertical angle of 6o° ; 
both cones, of course, having their 
vertices at O. Hence, when a and 
/? are given, we know that the line 
representing the force is an element of both cones ; and this is 
all that is given. 

(a) Now, if the sum of the two given angles is less than 
90 , the cones will not intersect, and the data are consequently 
inconsistent. 




DETERMINATION OF THE THIRD ANGLE. 



27 



(b) If, on the other hand, one of the given angles being 
greater than 90 , their difference is greater than 90 , the cones 
will not intersect, and the data are again inconsistent. 

(c) If a -f- ft = 90 , the cones are tangent to each other, 
and 7 = 90 . 

(d) If a + j3 > 90 , and a — /? or /3 — a < 90°, the cones 
intersect, and have two elements in common ; and we have no 
means of determining, without more. data, which intersection 
is intended, this being the indetermination that arises in the 
algebraic solution. 



EXAMPLES. 

[F = 6 3 a = 53 /? = 4 2° 

1. Given i F x = 49 a = 8f 7 = 72 

[f 2 = 2 fi = 70 7 = 45° 



Find the magnitude 
and direction of 
the resultant. 



Solution. 



F 


a. 


p. 


Y- 


COS a. 


COS/3. 


COSy. 


F COS a. 


F COS/3. 


F cos 7. 


63 

49 

2 


53° 
87 


42° 
70 


72° 

45° 


O.60182 
O.05234 
O.61888 


O.743I4 
O.94961 
O.34202 


O.2925O 
O.30902 
O.707 1 1 


37.91466 
2.56466 
I.23776 


46.81782 

46.53089 

0.68404 


18.42750 
15.14198 
1.41422 














4I.71708 

ZFcos a 


94-03275 

s^cos a 


34-98370 
2.E cos y 



R = \I(%F cos a) 2 + (S^cosyQ) 2 + (XFcos 7 ) 2 = 108.6569. 

log 2^cosa = 1.620314 log "EFcosfi = 1.973279 log 2/^ cosy = 1.543866 
logi? =2.036057 \ogR =2.036057 logi? =2.036057 



log cos a r =9.584257 log cos (3 r =9.937222 log cos y r =9.507809 
a r •= 67 25' 20" j3 r = 30° 4' 14" » 



28 



APPLIED MECHANICS. 



2. 



F. 



4-3 

87.5 

6.4 



47° *' 



88° 
68° 



65° i 

io° 5' 

8 3 ° 2' 





.P. 


a. 


p. 


V- 


3- 


5 
7 
4 


90° 

o° 


90° 






75 


73° 




45° 



§ 27. Conditions of Equilibrium for Forces applied at a 
Single Point. 

i°. When the forces are not confined to one plane, we have 
already found, for the square of the resultant, 

R = (XFcosa) 2 + (S^cos/3) 2 + (ZFcosy) 2 . 

But this expression can reduce to zero only when we have 

X^cosa = o, 'XFcosfi = o, and 2F cos y = o; 

for the three terms, being squares, are all positive quantities, 
and hence their sum can reduce to zero only when they are 
separately equal to zero. 

Hence : If a set of balanced forces applied at a single point 
be resolved into components along three directions at right angles 
to each other, the algebraic sum of the components of the forces 
along each of the three directions must be equal to zero, and con- 
versely. 

2°. When the forces are all confined to one plane, let that 
plane be the z plane ; then y — 90 in each case, and 

.-. p = 9 o° - a 

.*. cos/3 = sin a 

.-. R 2 = (ZFcosa) 2 4- (XFsina) 2 . 

Hence, for equilibrium we must have 

(2,F cos a) 2 + (XFsina) 2 = o; 



STATICS OF RIGID BODIES. 29 

and, since this is the sum of two squares, 

S^cosa = o, and *HF sin a = o. 

Hence : If a set of balanced forces, all situated in one plane, 
a7id acting at one point, be resolved into components along tzuo 
directions at right angles to each other, and in tJieir own plane, 
the algebraic sum of the components along each of the tzuo given 
directions must be equal to zero respectively; and conversely. 

§ 28. Statics of Rigid Bodies. — A rigid body is one that 
does not undergo any alteration of shape when subjected to 
the action of external forces. Strictly speaking, no body is 
absolutely rigid ; but different bodies possess a greater or less 
degree of rigidity according to the material of which they are 
composed, and to other circumstances. When a force is ap- 
plied to a rigid body, we may have as the result, not merely a 
rectilinear motion in the direction of the force, but, as will be 
shown later, this may be combined with a rotary motion ; in 
short, the criterion by which we determine the ensuing motion 
is, that the effect of the force will distribute itself through the 
body in such a way as not to interfere with its rigidity. 

What this mode of distribution is, we shall discuss here- 
after ; but we shall first proceed to some propositions which can 
be proved independently of this consideration. 

§ 29. Principle of Rectilinear Transferrence of Force in 
Rigid Bodies. — If a force be applied to a rigid body at the 
point A (Fig. 13) in the direction AB, 
whatever be the motion that this force 
would produce, it will be prevented from 
taking place if an equal and opposite 
force be applied at A, B, C, or D, or at FlG - x 3- 

any point along the line of action of the force : hence we have 
the principle that — 

TJie point of application of a force acting on a rigid body, 
may be transferred to any other point which lies in the line of 




30 



APPLIED MECHANICS. 



action of the force, and also in the body, without altering the 
resulting motion of the body, although it does alter its state of 
stress. 

§ 30. Composition of two Forces in a Plane acting at 
Different Points of a Rigid Body, and not Parallel to Each 
Other. — Suppose the force F (Fig. 14) to be applied at A, and 
Fj at By both in the plane of the paper, and acting on the rigid 
body abcdef Produce the lines of direction of the forces till 
they meet at O, and suppose both F and F t to act at O. Con- 
struct the parallelogram ODHE, where OD = F and OF = F x ; 

then will OH — R rep- 
resent the resultant 
force in magnitude and 
in direction. Its point 
d of application may be 
conceived at any point 
along the line OH, as 
at C, or any other 
point ; and a force 
equal and opposite to 
OH, applied at any point of the line OH, will balance F at A, 
and F, at B. 

The above reasoning has assumed the points A, B, C and 
O, all within the body : but, since we have shown, that when 
this is the case, a force equal and opposite to R at C will bal- 
ance Fat A, and F x at B, it follows, that were these three forces 
applied, equilibrium would still subsist if we were to remove 
the part bafegho of the rigid body ; or, in other words, — 

The same construction holds even wJien the point O falls out- 
side the rigid body. 

§31. Moment of a Force with Respect to an Axis Per- 
pendicular to the Force. 

Definition. — The moment of a force with respect to an 
axis perpendicular to the force, and not intersecting it, is the 




EQUILIBRIUM OF THREE PARALLEL FORCES. 



31 




Fig. 15. 



product of the force by the common perpendicular to (shortest 
distance between) the force and the axis. 

Thus, in Fig. 15 the moment of F about 
an axis through O and perpendicular to the 
plane of the paper is F{OA). The sign of 
the moment will depend on the sign attached 
to the force and that attached to the perpen- 
dicular. These will be assumed in this book 
in such a manner as to render the following true ; viz,, — 

The moment of a force zvith respect to an axis is called posi- 
tive when, if the axis were supposed fixed, the foixe would cause 
the body on which it acts to rotate around the axis in the direc- 
tion of the hands of a watch as 
seen by the observer looking at 
the face. It will be called nega- 
tive when the rotation would take 
place in the opposite direction. 

§ 32. Equilibrium of Three 
Parallel Forces applied at 
Different Points of a Rigid 
Body. — Let it be required to 
find a force (Fig. 16) that will 
balance the two forces F at A, 
and F, at B. Apply at A and B 
respectively, and in the line AB, 
the equal and opposite forces Aa 
and Bb. .Their introduction will 
produce no alteration in the 
body's motion. 

The resultant of F and Aa 

is Afi that of F, and Bb is Bg. 

Compound these by the method 

of § 30, and we obtain as result- 

A force equal in magnitude and opposite in direction 




Fig. 16. 



ant ce. 



32 



APPLIED MECHANICS. 



to ce, applied at any point of the line cC, will be the force 
required to balance F 2X A and F t at B ; and, as is evident from 
the construction, this line is in the plane of the two forces. 
Moreover, by drawing triangle fKl equal to Bbg, we can readily 
prove that triangles oce and Afl are equal : hence the angle oce 
equals the angle fAl, and R is parallel to .Fand F t . Also 

R = ce = ch + he = AK + Kl = F + F l9 



and 







Cc __AK _ 7? 
^C /A" ^« 






1 




G- .. & __ jP, . 

BC Bb Bb' 






since Aa 


= Bb, 








AC _ F, 
BC F 


F 
' BC 


F t _F+ F t : F _ 

AC AB Cr 


F f 
Cq~ 


F + F x 
qr 



where qr is any line passing through C. 

Hence we have the following propositions ; viz., — 
If three parallel forces balance each other, — 
i°. They must lie in one plane. 

2°. The middle one mnst be equal in magnitude and opposite 
F in direction to tlie sum of the other 

two. 

3°. Each force is proportional to the 
distance between the lines of direction 
of tJie other two as - measured on any 
line intersecting all of them. 

The third of the above-stated con- 
ditions may be otherwise expressed, 
thus : — 

The algebraic sum of the moments 
of the three forces about any axis perpendicular to the forces 
must be zero. 



f 



Fig. 17. 



RESULTANT OF A PAIR OF PARALLEL FORCES. 33 

Proof. — Let F, F lt and R (Fig. 17) be the forces ; and let 
the axis referred to pass through O. Draw OA perpendicular 
to the forces. Then we have 

F(OA) + F t (OB) = F(OC + CA) + FiOC - BC) 

= (F+ F s )OC + F(AC) - FABC). 

But, from what we have already seen, 

F+F t = -R 
and 

BC AC 
.-. F(AC) = F t {BC) 
.'. F(OA) + FiOB) = -F(OC) + o 
.'. F{OA) + F,(OB) + R{OC) = o, 

or the algebraic sum of the moments of the forces about the 
axis through is equal to zero. 

§ 33. Resultant of a Pair of Parallel Forces. — In the 
preceding case, the resultant of any two of the three forces 
F y F n and F, in Fig. 16 or Fig. 17, is equal and opposite to the 
third force. Hence follow the two propositions : — 

I. If two parallel forces act in the same, direction, their 
resultant lies in the plane of the forces, is equal to their sum,. 
acts in the same direction, and cuts the line joining their points 
of application, or any common perpendicular to the two forces, 
at a point which divides it internally into two segments in- 
versely as the forces. 

II. If two unequal parallel forces act in opposite directions, 
their resultant lies in the plane of the forces, is equal to their 
difference, acts in the direction of the larger foixe, and cuts the 
line joining their points of application, or any common perpen- 
dicular to them, at a point which (lying nearer the larger force) 



34 APPLIED MECHANICS. 

divides it externally into two segments which are inversely as 
the forces. 

Another mode of stating the above is as follows : — 

i°. The resultant of a pair of parallel forces lies in the plane 
of the forces. 

2°. It is equal in magnitude to their algebraic sum, and coin- 
cides in direction with the larger force. 

3°. The moment of the resultant about an axis perpendicu- 
lar to the plane of the forces is equal to the algebraic sum of 
the moments about the same axis. 

EXAMPLES. 

i. Find the length of each arm of a balance such that i ounce at 
the end of the long arm shall balance i pound at the end of the short 
arm, the length of beam being 2 feet, and the balance being so propor- 
tioned as to hang horizontally when unloaded. 

2. Given beam =28 inches, 3 ounces to balance 15. 

3. Given beani = 36 inches, 5 ounces to balance 25 ounces. 



MODE OF DETERMINING THE RESULTANT OF A PAIR OF PARALLEL 
FORCES REFERRED TO A SYSTEM OF THREE RECTANGULAR 
AXES. 

Let both forces (Fig. 18) be parallel to OZ ; then we have, 
from what has preceded, 

Z = £1 = Z±Ii = * W here Rm.l?-+-JP» 

be ab ac ae 

But from the figure 

be = ab . F _ F x 

CXrn ~— 1X2 «%2 *^I PC* "— ~ DC 2 *™2 " "" " *^I 

Fx 2 — Fx t = F v x z — F t x 2 

/. (F+ F s )x 2 = Fx, + F T x 2 , 
or 

Rx 2 = Fx t 4- FtX 3 ; 



RESULTANT OF NUMBER OF PARALLEL FORCES. 



35 



and similarly we may prove that 

Ry 2 = Fy x + F,y z , 



or 



i°. The resultant of two parallel forces is parallel to the 
forces and equal to their algebraic sum. 



aR=F+F, 




Fig. iS. 

2°. The moment of the resultant with respect to OX is 
equal to the algebraic sum of their moments with respect to 
OX; and likewise when the moments are taken with respect 
to OY. 

§ 34. Resultant of any Number of Parallel Forces. — 
Let it be required to find the resultant of any number of paral- 
lel forces. 

In any such case, we might begin by compounding two of 
them, and then compounding the resultant of these two with a 
third, this new resultant with a fourth, and so on. Hence, for 
the magnitude of any one of these resultants, we simply add 
to the preceding resultant another one of the forces ; and for 
the moment about any axis perpendicular to the forces, we add 



36 



A PPL TED MECHANICS. 



to the moment of the preceding resultant the moment of the 
new force. 

Hence we have the following facts in regard to the resultant 
of the entire system : — 

I °. The resultant will be parallel to the forces and equal to 
their algebraic sum. 

2°. The moment of the resultant about any axis perpendicular 
to the forces will be equal to the algebraic s?tm of the moments 
of the forces about the same axis. 

The above principles enable us to determine the resultant 
in all cases, except when the algebraic sum of the forces is 
equal to zero. This case will be considered later. 

§ 35. Composition of any System of Parallel Forces 



f 3 f, 



Fig. 19. 



when all are in One Plane. — 
Refer the forces to a pair of rect- 
angular axes, OX, OY (Fig. 19), 
and assume OY parallel to the 
forces. 

The forces and the co-ordinates 
of their lines of direction being as 
indicated in the figure, if we denote 
by R the resultant, and by x Q the 
co-ordinate of its line of direction, 
we shall have, from the preceding, 

R = XF; (1) 

and if moments be taken about an 
axis through O, and perpendicular 



to the plane of the forces, we shall also have 

Rx Q = "ZFx. 
Hence 

R = 2,F and x Q = 



(») 



%Fx 



%F 



determine the resultant in magnitude and in line of action, 
except when ^F = o, which case will be considered later. 



EQUILIBRIUM OF ANY SET OF PARALLEL FORCES. 2>7 

§ 36. Composition of any System of Parallel Forces not 
confined to One Plane. — Refer the forces to a set of rect- 
angular axes so chosen that OZ is parallel to their direction. 
If we denote the forces by F n F 2 , F z , F 4 , etc., and the co-ordinates 
of their lines of direction by (,r„ y s ), (x 2 , y 2 ), etc., and if we 
denote their resultant by R, and the co-ordinates of its line of 
direction by (x oi y a ), we shall have, in accordance with what has 
been proved in § 34, — 

i°. The magnitude of the resultant about OY is equal to tlie 
algebraic sum of the forces, or 

R = S/f. 

2°. The moment of the resultant about OY is equal to the 
sum of the moments of the forces about OY, or 

x ^F = %Fx. 

3 . The moment of tlie resultant about OX is equal to the 
sum of the moments about OX, or 

y Q ^F = LFy. 



Hence 



~ p %Fx IFy 

*F, *o=— , *=YF' 



determine the resultant in all cases, except when ^F =. o. 

§ 37. Conditions of Equilibrium of any Set of Parallel 
Forces. — If the axes be assumed as before, so that OZ is 
parallel to the forces, we must have 

%F = o, 2Fx = o, and ^.Fy = o. 

To prove this, compound all but one of the forces. Then equilib- 
rium will subsist only when the resultant thus obtained is equal 
and directly opposed to the remaining force ; i.e, it must be 
equal, and act along the same line and in the opposite direction. 
Hence, calling R a the resultant above referred to, and (x m y a ) 
the co-ordinates of its line of direction, and calling F M the 



38 APPLIED MECHANICS. 

remaining force, and (x n , y,) the co-ordinates of its line of direc- 
tion, we must have 

Ra = —F n , X a = X„, y a = J>n, 

.'. R a + F„= o, R a x a -h FnX n — o, R a y a + F n y n = o, 
.'. ^F = o, ^Fx = o, %Fy = o. 

When the forces are all in one plane, the conditions become 

5F = o, %Fx = o. 

§ 38. Centre of a System of Parallel Forces, — The 

resultant of the two parallel forces F and F l (Fig. 20), ap- 
plied at A and B respectively, is a force R = F + F lf whose 
line of action cuts the line AB at a point C, 
which divides it into two segments inversely as 
the forces. If the forces F and F t are turned 
through the same angle, and assume the posi- 
tions AO and BO l respectively, the line of 
action of the resultant will still pass through 
C, which is called the centre of the two parallel 
forces F and F t . Inasmuch as a similar reasoning will apply 
in the case of any number of parallel forces, we may give the 
following definition : — 

The centre of a system of parallel forces is the point through 
which the line of action of the resultant always passes, no matter 
how the forces are turned, provided only — 

1°. Their points of application remain the same. 
2°. Their relative rnagnitudes are unchanged. 
3°. They remain parallel to each other. 

Hence, in finding the centre of a set of parallel forces, we 
may suppose the forces turned through any angle whatever, and 
the centre of the set is the point through which the line of 
action of the resultant always passes. 




DISTRIBUTED FORCES. 



39 



§ 39. Co-ordinates of the Centre of a Set of Parallel 
Forces. — Let F x (Fig. 21) be one of the forces, and {x xy y xx z x ) 
the co-ordinates of its point 



of application. Let F 2 be 
another, and (x 2 , y 2 , z 2 ) co- 
ordinates of its point of 
application. Turn all the 
forces around till they are 
parallel to OZ, and find the 
line of direction of the re- 
sultant force when they are 
in this position. The co- 
ordinates of this line are 




XFx 



Fig. 2i. 



y* 



IFy . 



and, since the centre of the system is a point on this line, the 
above are two of the co-ordinates of the centre. Then turn 
the forces parallel to OX, and determine the line of actum of 
the resultant. We shall have for its co-ordinates 



y ~^y 



2J% 
2F' 



Hence, for the co-ordinates of the centre of the system, we 
have 



2Fx 
~2F' 



}o " 2F' 



2Fz 
2F' 



When %F = o the co-ordinates would be 00, therefore such 
a system has no centre. 

§ 40. Distributed Forces While we have thus far as- 
sumed our forces as acting at single points, no force really acts 
at a single point, but all are distributed over a certain surface 



40 APPLIED MECHANICS. 

or through a certain volume ; nevertheless, the propositions 
already, proved are all applicable to the resultants of these 
distributed forces. We shall proceed to discuss distributed 
forces only when all the elements of the distributed force are 
parallel to each other. As a very important example of such a 
distributed force, we may mention the force of gravity which 
is distributed through the mass of the body on which it acts. 
Thus, the weight of a body is the resultant of the weights of 
the separate parts or particles of which it is composed. As 
another example we have the following : if a straight rod be 
subjected to a direct pull in the direction of its length, and if 
it be conceived to be divided into two parts by a plane cross- 
section, the stress acting at this section is distributed over the 
surface of the section. 

§41. Intensity of a Distributed Force. — Whenever we 
have a force uniformly distributed over a certain area, we obtain 
its intensity by dividing its total amount by the area over which 
it acts, thus obtaining the amount per unit of area. 

If the force be not uniformly distributed, or if the intensity 
vary at different points, we must adopt the following means 
for finding its intensity. Assume a small area containing the 
point under consideration, and divide the total amount of force 
that acts on this small area by the area, thus obtaining the 
mean intensity over this small area: this will be an approxima- 
tion to the intensity at the given point ; and the intensity is the 
limit of the ratio obtained by making the division, as the area 
used becomes smaller and smaller. 

Thus, also, the intensity, at a given point, of a force which 
is distributed through a certain volume, is the limit of the 
ratio of the force acting on a small volume containing the 
given point, to the volume, as the latter becomes smaller and 
smaller. 

§42. Resultant of a Distributed Force. — i°. Let the 
force be distributed over the straight line AB (Fig. 22), and 



RESULTANT OF A DISTRIBUTED FORCE. 



41 




let its intensity at the point E where AE = x, be represented 
by EF ' = / = <j>(x), a function of x ; 
then will the force acting on the por- 
tion Ee — Ax of the line be/A^r; and 
if we denote by R the magnitude of 
the resultant of the force acting on the 
entire line AB, and by x Q the distance 
of its point of application from A, we shall have 

R = %p&x approximately, 
or 

R = fpdx exactly ; 

and, by taking moments about an axis through A perpendicular 
to the plane of the force, we shall have 

x Q R — ^x(pAx) approximately, 
or 

x Q R = fpxdx exactly ; 

whence we have the equations 

fpxdx 
R = fpdx, x a = -— 

2°. Let the force be distributed over a plane area EFGH 

(Fig. 23), let this area be re- 




ferred to a pair of rectangular 
axes OX and Q Y, in its own 
plane, and let the intensity 
of the force per unit of area 
at the point P, whose co- 
ordinates are x and y, be 
p = <f>(x, y) ; then will pXrAy 
be approximately the force act- 
ing on the small rectangular 
area Xr&y. Then, if we rep- 
resent by R the magnitude of 
the resultant of the distributed force, and by x , y Q , the co-ordi- 



42 APPLIED MECHANICS. 

nates of the point at which the line of action of the resultant 
cuts the plane of EFGH, we shall have 

R — 'ZpAxAy approximately, 
x R = 2x(p&xAy) " 
y R = Zy(pAxAy) " 

or, as exact equations, we shall have 

R = ffpdxdy, 

_ ffpxdxdy ffpydxdy 

X ° ~ Ifpdxdy ' y ° ~ ffpdxdy ' 

3°. Let the force be distributed through a volume, let this 
volume be referred to a system of rectangular axes, OX, OY y 
and OZ, let A V represent the elementary volume, whose co- 
ordinates are x, y, z, and let p = 4>(x, y, z) be the intensity of 
the force per unit of volume at the point (x, y, z) ; then, if we 
represent by R the magnitude of the resultant, and by x , j/ oy z ot 
the co-ordinates of the centre of the distributed force, we shall 
have, from the principles explained in § 38 and § 39, the approx- 
imate equations 

R = S/A V, x R = ^x{p\ V), y Q R = 2j'(/A V), Zo R= $z(p± V) ; 
and these give, on passing to the limit, the exact equations 

R -rp d v X -IMI v -iMZ z -iMI 

§ 43. Centre of Gravity. — The weight of a body, or system 
of bodies, is the resultant of the weight of the separate parts 
or particles into which it may be conceived to be divided ; and 
the centre of gravity of the body, or system of bodies, is the 
centre of the above-stated system of parallel forces, i.e., the 
point through which the resultant always passes, no matter how 
the forces are turned. The weight of any one particle is the 
force which gravity exerts on that particle : hence, if we repre- 



FORCE APPLIED TO CENTRE OF STRAIGHT ROD. 43 

sent the weight per unit of volume of a body, whether it be 
the same for all parts or not, by w, we shall have, as an 
approximation, 

W-^w^V, *°-,^ AF > A-.fcAjr' z °- %w ±v' 

and as exact equations, 

fwxdV fwydV fwzdV 

W^fwdV, x^j^y, y°=J^fv> *°=ywF> (0 

where £F denotes the entire weight of the body, and x , y m z of 
the co-ordinates of its centre of gravity. 

If, on the other hand, we let M =: entire mass of the body, 
dM = mass of volume dV, and m — mass of unit of volume, 
we shall have 

W == Mg, w = mg, wdV =mgdV = gdM. 

Hence the above equations reduce to 

mr rsiur f xdM SydM fzdM . . 

M=fdM, *o = W , yo^-jr^, *o=-jr^. (2) 

Equations (1) and (2) are both suitable for determining the 
centre of gravity; one of the sets being sometimes most con- 
venient, and sometimes the other. 

§44. Centre of Gravity of Homogeneous Bodies If 

the body whose centre of gravity we are seeking is homogeneous, 
or of the same weight per unit of volume throughout, we shall 
have, that w — a constant in equations (1) ; and hence these 
reduce to 

§45. Effect of a Single Force applied at the Centre of 
a Straight Rod of Uniform Section and Material. — If a 
straight rod of uniform section and material have imparted to it 




44 APPLIED MECHANICS. 

a motion, such that the velocity imparted in a unit of time to 
each particle of the rod is the same, and if we represent this 
velocity by/, then if at each point of the rod, we lay off a line 
xy (Fig. 24) in the direction of the motion, 
and representing the velocity imparted to that 
point, the line bounding the other ends of 
the lines xy will be straight, and parallel to the 
rod. If we conceive the rod to be divided 
into any number of small equal parts, and 
denote the mass of one of- these parts by \M, then will f\M 
contain as many units of momentum as there are units of force 
in the force required to impart to this particle the velocity 
f in a unit of time ; and hence f±M is the measure of this 
force. 

Hence the resultant of the forces which impart the velocity 
f to every particle of the rod will have for its measure 

fM t 

where M is the entire mass of the rod ; and its point of applica- 
tion will evidently be at the middle of the rod. 

It therefore follows that — 

The effect of a single fore -e applied at the middle of a straight 
rod of uniform section and material is to impart to the rod a 
motion of translation in the direction of the force, all points of 
the rod acquiring equal velocities in equal times. 

§46. Translation and Rotation combined. — Suppose that 
we have a straight rod AB (Fig. 25), and suppose that such a 
force or such forces are applied to it as will impart to the point 
A in a unit of time the velocity Aa, and to the point B the 
(different) velocity />£ in a unit of time, both being perpendicu- 
lar to the length of the rod. It is required to determine the 
motion of any other point of the rod and that of the entire 
rod. 



TRANSLATION AND ROTATION COMBINED. 



45 




Fig. 25. 



Lay off Aa and Bb (Fig. 25), and draw the line ab, and pro- 
duce it till it meets AB produced 
in 0: then, when these velocities 
Aa and Bb are imparted to the 
points A and B, the rod is in the 
act of rotating around an axis 
through perpendicular to the plane of the paper ; for when a 
body is rotating around an axis, the linear velocity of any point 
of the body is perpendicular to the line joining the point in 
question with the axis (i.e., the perpendicular dropped from the 
point in question upon the axis), and proportional to the dis- 
tance of the point from the axis. 

Hence : If the velocities of two of the points in the rod are 
given, and if these are perpendicular to the rod, the motion 
of tlie rod is fixed, and consists of a rotation about some axis 
at right angles to the rod. 

Another way of considering this motion is as follows : Sup- 
pose, as before, the velocities of the points A and B to be 





A 


?2 


V? 


**"^«^ 


a 


a 


X 

V 

h 6 


""""■■ ~~» 



represented by Aa and Bb respec- 
tively, and hence the velocity of 
any other point, as x (Fig. 26), to 
be represented by xy, or the length 
of the line drawn perpendicular to 
AB, and limited by AB and ab. 
Bb, — \{Aa -f- Bb) — Cc, and draw 



Fig. 26. 

Then, if we lay off Aa, 

a,b„ and if we also lay off Aa 2 = a,a, and Bb 2 = bj), we shall 

have the following relations ; viz., — 



Aa = Aa, — Aa 2 , 
Bb = Bb, + Bb 2 , 
xy = xy, — xy 2 , 



etc., 



or we may say that the actual motion imparted to the rod in a 
unit of time may be considered to consist of the following two 
parts : — 



46 APPLIED MECHANICS. 

i°. A velocity of translation represented by Aa lf the mean 
velocity of the rod ; all points moving with this velocity. 

2°. A varying velocity, different for every different point, 
and such that its amount is proportional to its distance from 
C, the centre of the rod, as graphically shown in the triangles 
Aa 2 CBb 2 . In other words, the rod has imparted to it two 
motions : — 

i°. A translation with the mean velocity of the rod. 

2°. A rotation of the rod about its centre. 

§ 47. Effect of a Force applied to a Straight Rod of 
Uniform Section and Material, not at its Centre. — If the 
force be not at right angles to the rod, resolve it into two com- 
ponents, one acting along the rod, and the other at right angles 
to it. The first component evidently produces merely a trans- 
lation of the rod in the direction of its length : hence the second 
component is the only one whose effect we need to study. 

To do this we shall proceed to show, that, when such a rod 
has imparted to it the motion described in § 46, the single re- 
sultant force which is required to impart 
this motion in a unit of time is a force 
acting at right angles to the rod, at a point 
different from its centre ; and we shall de- 
fig. 27. termine the relation between the force and 

the motion imparted, so that one may be deduced from the 
other. 

Let A be the origin (Fig. 27), and let 

Ac = x, cd = dx. 

AB — I — length of the rod. 

ce =/= velocity imparted per unit of time at distance x 
from A. 

Aa =/„ Bb =/_ 

w — weight per unit of length. 

71) 

in = mass per unit of length = — . 

S 




EFFECT OF FORCE APPLIED TO A STRAIGHT ROD. 4? 

W = entire weight of rod. 

W 
M ■=■ entire mass of rod — — . 

S 
R = single resultant force acting for a unit of time to 

produce the motion. 
x Q = distance from A to point of application of R. 

Then we shall have, 

Hence, from § 42, 

] fdx = **(area AabB) = -(/ 4-/ 2 )/ = -(/, +/ 2 ). (1) 

o 2 2 

X R = «J>^ = J (/ + 2/,)/" = £-(/ + 2/ 2 )/. (2) 

. „ i '/i •+ 2/ 2 , 



3 /, +/ 2 



(3) 



We thus have a force R> perpendicular to AB, whose mag- 
nitude is given by equation (1), and whose point of application 
is given by equation (3) ; the respective velocities imparted by 
the force being shown graphically in Fig. 27. 



EXAMPLES. 

1. Given Weight of rod = W = 100 lbs., 
Length of rod = 3 feet, 

Assume £■ = 32 feet per second, 

Force applied = R = 5 lbs., 
Point of application to be 2.5 feet from one end; 

determine the motion imparted to the rod by the action of the force for 
one second. 



4<3 APPLIED MECHANICS, 



Solution. 
Equation (i) gives us, 

/roo\/ f +/ 2 

Equation (2) gives, 

(2.5) (5) = ('"y) © (3)(/x + 2/ 2 ), or/ + 2 / 2 = 8 

.-. / 2 = 4-8, / = -i.e. 

Hence the rod at the end nearest the force acquires a velocity of 4.8 
feet per second, and at the other end a velocity of —1.6 feet per 
second. The mean velocity is, therefore, 1.6 feet per second; and we 
may consider the rod as having a motion of translation in the direc- 
tion of the force with a velocity of 1.6 feet per second, and a rotation 
about its centre with such a speed that the extreme end (i.e., a point 
§ feet from the centre) moves at a velocity 4.8 — 1.6 = 3.2 feet per 

second. Hence angular velocity = — = 2.14 per second = i22°.6 

per second. 

2. Given W = 50 lbs., / = 5 feet. It is desired to impart to it, 
in one second, a velocity of translation at right angles to its length, of 5 
feet per second, together with a rotation of 4 turns per second : find the 
force required, and its point of application. 

3. Assume in example 2 that the velocity of translation is in a 
direction inclined 45 ° to the length of the rod, instead of 90 . Solve 
the problem. 

4. Given a force of 3 lbs. acting for one-half a second at a distance 
of 4 feet from one end of the rod, and inclined at 30 to the rod : 
determine its motion. 

5. Given the same conditions as in example 4, and also a force 
of 4 lbs., parallel and opposite in direction to the 3-lb. force, and acting 
also for one-half a second, and applied at 3 feet from the other end : 
determine the resulting motion. 



MOMENT OF THE FORCES CAUSING ROTATION 49 

6. Given two equal and opposite parallel forces, each acting at right 
angles to the length of the rod, and each equal to 4 lbs., one being 
applied at 1 foot from one end, and the other at the middle of the rod ; 
find the motion imparted to the rod through the joint action of these 
forces for one-third of a second. 

§ 48. Moment of the Forces causing Rotation. — Re- 
ferring again to Fig. 26, and considering the. motion of the 
rod as a combination of translation and rotation, if we take 
moments about the centre C, and compare the total moment 
of the forces causing the rotation alone, whose accelerations 
are represented by the triangles aa^cbj?, with the total moment 
of the actual forces acting, whose accelerations are represented 
by the trapezoid AabB, we shall find these moments equal to 
each other ; for, as far as the forces represented by the rectangle 
are concerned, every elementary force m(xy 1 )dx on one side of 
the centre C has its moment (Cx)\m(xy I )dx\ equal and opposite 
to that of the elementary force at the same distance on the other 
side of C: hence the total moment of the forces represented 
graphically by the rectangle AaJ) x B is zero, and hence — 

The moment about C of those represented by the trapezoid 
equals the moment of those represented by tlie triangles. 

Hence, from the. preceding, and from what has been pre- 
viously proved, we may draw the following conclusions : — 

i°. If a force be applied at the centre of the rod, it will 
impart the same velocity to each particle. 

2°. If a force be applied at a point different from the centre, 
and act at right angles to its length, it will cause a translation 
of the rod-, together with a rotation about the centre of the rod. 

3 . In this latter case, the moment of the forces imparting 
the rotation alone is equal to the moment of the single resultant 
force about the centre of the rod, and the velocity of translation 
imparted in a unit of time is equal to the number of units of 
force in the force, divided by the entire mass of the rod. 



5o 



APPLIED MECHANICS. 



§ 49. Effect of a Pair of Equal and Opposite Parallel 
Forces applied to a Straight Rod of Uniform Section and 
Material. — Suppose the rod to be AB (Fig. 28), and let the 
two equal and opposite parallel forces be Dd and Ee, each equal 
to F, applied at D and E respectively. 
The mean velocity imparted in a unit 

of time by either force will be — ; and, 

from what we have already seen, the trap- 
ezoid AabB will furnish us the means of 
representing the actual velocity imparted 
to any point of the rod by the force Dd. 
The relative magnitudes of Aa and Bb, the 
accelerations at the ends, will depend, of 

course, on the position of D ; but we shall 

p 
always have Cc — \{Aa -f- Bb) = — , a 

quantity depending only on the magnitude 

of the force. So, likewise, the trapezoid Aa,b t B will represent 

the velocities imparted by the force Ee ; and while the relative 

magnitude of Aa t and Bb x will depend upon the position of E } 

27 
we shall always have Cc, = \{Aa x + Bb) = — . Hence, since 

Cc = Cc lf the centre C of the rod has no motion imparted to it 
by the given pair of forces, hence the motion of the rod is one 
of rotation about its centre C. 

The resulting velocity of any point of the rod will be the 
difference between the velocities imparted by the two forces ; 
and if these be laid off to scale, we shall have the second 
figure. Hence — 

A pair of equal and opposite parallel foixes, applied to a 
straight rod of uniform section and material, produce a rota- 
tio7i of the rod about its centre. Also, — 

Such a rotation about the centre of the rod cannot be pro- 




Fig. 28. 



EFFECT OF STATICAL COUPLE ON STRAIGHT ROD. 5 I 



duced by a single force, but requires a pair of equal and op- 
posite parallel forces. 

§ 50. Statical Couple. — A pair of equal and opposite 
parallel forces is called a statical couple. 

§51. Effect of a Single Force applied at the Centre of 
Gravity of a Straight Rod of Non-Uniform Section and 
Material. — In the case of a straight rod of non-uniform sec- 
tion and material, we may consider the rod as composed of a 
set of particles of unequal mass : and if we imagine each par- 
ticle to have imparted to it the same velocity in a unit of time, 
then, using the same method of graphical representation as 
before (Fig. 24), the line ab, bounding the other ends of the 
lines representing velocities, will be parallel to AB ; but if we 
were to represent by the lines xy, not the velocities imparted, 
but the forces per unit of length, the line bounding the other 
ends of these forces would not, in this case, be parallel to AB. 
Moreover, since these forces are proportional to the masses, and 
hence to the weights of the several particles, their resultant 
would act at the centre of gravity of the rod. Hence — 

A force applied at the centre of gravity of a straight rod will 
impart the same velocity to each poirit of the rod ; i.e., will im- 
part to it a motion of translation only. 

§ 52. Effect of a Statical Couple on a Straight Rod of 
Non-Uniform Section and Material. — Let such a rod have 
imparted to it only a motion of rotation about its centre of 
gravity, and let us adopt the same modes of graphical repre- 
sentation as before. 

Let the origin be taken at O (Fig. 29), 
the centre of gravity of the rod. 

Let Aa = f = velocity imparted to A. 
Bb = f 2 = velocity imparted to B. 
OA = a, OB = b, OC = x, 
CD = f = velocity imparted to C. 
dM = elementary mass at C. 




$2 APPLIED MECHANICS. 

Then, from similar triangles, we have 

a b 

and hence for the force acting on dM we have 

dF = (CF)dx = ^xdM. 
a 

Hence the whole force acting on AO, and represented graph- 
ically by Aa.O, is 

*- C*xdM, 



and that acting on OB, and represented by BOb x , is 

Jf*x = o / f*x = o 

xdM = J -l I xdM. 
x = -b ajx = -b 



b, 



Hence for the resultant, or the algebraic sum, of the two, we 
have 

-ff = - f X xdM. 

a Jx = -b 

But from § 43 we have for the co-ordinate x of the centre of 
gravity of the rod 

JOx = a 
xdM 
* = -* . 

°~ ■ M ' 

but, since the origin is at the centre of gravity, we have 

x n = o, 



and hence 



x 



x = a 

xdM — o .*. R = o. 

= -b 



Hence the two forces represented by Aa x O and B3 x O are equal 
in magnitude and opposite in direction : hence the rotation 
about the centre of gravity is produced by a Statical Couple. 



MEASURE OF THE ROTATORY EFFECT. 53 

Now, a train of reasoning similar to that adopted in the case 
of a rod of uniform section and material will show that a single 
force applied at some point which is not the centre of gravity 
of the rod will produce a motion which consists of two parts ; 
viz., a motion of translation, where all points of the rod have 
equal velocities, and a motion of rotation around the centre of 
gravity of the rod. 

§53. Moment of a Couple. — The moment of a statical 
couple is the product of either force by the perpendicular dis- 
tance between the two forces, this perpendicular distance being 
called the arm of the couple. 

§ 54. Measure of the Rotatory Effect. — Before proceed- 
ing to examine the effect of a statical couple upon any rigid 
body whatever, we will seek a means of measuring its effect in 
the cases already considered. 

The measure adopted is the moment of the couple ; and, in 
order to show that it is proper to adopt this measure, it will be 
necessary to show — 

That the moment of the couple is proportional to the angu- 
lar velocity imparted to the same rod in a unit of time ; and 
from this it will follow — 

That two couples in the same plane with equal moments will 
balance each other if one is right-handed and the other left-handed. 

If we assume the origin of co-ordinates at C (Fig. 30), the 
centre of gravity of the rod, and if we 
denote by a the angular velocity imparted 
in a unit of time by the forces F and — F, 
and let CD = x„ CE == x„ then we have 
for the linear velocity of a particle situated 
at a distance x from C the value 

ax. 

The force which will impart this velocity in a unit of time to 

the mass dM is 

axdM. 




54 APPLIED MECHANICS. 

The total resultant force is 

afxdM, 

which, as we have seen, is equal to zero. The moment of the 
elementary force about C is 

x(axdM) = ax 2 dM, 
and the sum of the moments for the whole rod is 

afx 2 dM, 

and this, as is evident if we take moments about C, is equal to 
Fx x - Fx 2 = F(x s — x z ) = F(DF). 

Now, fx-dM is a constant for the same rod : hence any quan- 
tity proportional to F{DE) is also proportional to a. 

The above proves the proposition. 

Moreover, we have 

F(DE) = afx*dM 
F{DF) 

.*. a == — -, 

fx-dM 

whence it follows, that when the moment of the couple is given, 
and also the rod, we can find the angular velocity imparted in 
a unit of time by dividing the former by fx 2 dM. 

§ 55. Effect of a Couple on a Straight Rod when the 
Forces are inclined to the Rod. — We shall next show that 
the effect of such a couple is the same as that of a couple of 
equal moment whose forces are perpen- 
dicular to the rod. 

In this case let AD and BC be the 
forces (Fig. 31). The moment of this 
couple is the product of AD by the per- 
pendicular distance between AD and BC, 

Fig 31 

the graphical representation of this being 
the area of the parallelogram ADBC. 



a^--*. 




EFFECT OF A STATICAL COUPLE ON A RIGID BODY. 55 

Resolve the two forces into components along and at right 
angles to the rod. The former have no effect upon the motion 
of the rod : the latter are the only ones that have any effect 
upon its motion. The moment of the couple which they form 
is the product of Ad by AB, graphically represented by paral- 
lelogram AdBb ; and we can readily show that 

ADBC = AdBb. 

Hence follows the proposition. 

§ 56. Effect of a Statical Couple on any Rigid Body. — 
Refer the body (Fig. 32) to three rectan- 
gular axes, OX, OY, and OZ, assuming 
the origin at the centre of gravity of the 
body, and OZ as the axis about which 
the body is rotating. Let the mass of the 
particle P be AM, and its co-ordinates be 
x, y, z. 

Then will the force that would impart fig. 32. 

to the mass AM the angular velocity a in a unit of time be 

arAM, 
where r = perpendicular from P on OZ, or 




r = ^x 2 -+- y 2 . 

This force may be resolved into two, one parallel to O Y and 
the other to OX; the first component being axAM, and the 
second ay AM. 

Proceeding in the same way with each particle, and finding 
the resultant of each of these two sets of parallel forces, we 
shall obtain, finally, a single force parallel to OY and equal to 

a^xAM, 
and another parallel to OX, equal to 

*%yAM. 



$6 APPLIED MECHANICS. 

But, since OZ passes through the centre of gravity of the body, 
we shall have 

"Zx&M = o and %y&M = o. 

Hence the resultant is in each case, not a single force, but a 
statical couple ; the moment of the first couple being 

a%x 2 AM, 
and that of the second 

aly*AM. 

These couples produce the same effect in whatever plane per- 
pendicular to OZ they are situated. Hence, suppose them both 
in the plane XOY, then representing them as in (Fig. 33), we 
make 

F,{AB) = a%x 2 AM and F 2 (CD) = aty^M. 

Now, compound the force at D with that at B, and the force at 

« c , C with that at A, and we obtain as a 

|f, result two equal and opposite parallel 

o |b x forces, or one statical couple, whose 



J 



F f" R ^ K moment will be shown to be equal 



to 



E F 2 



^G 



a($x 2 <\M + 2y 2 AM) 



a1(x 2 + y 2 )±M = aZrt&M. 
To show this, let 



Fig. 33. 



OB = p l = one-half the arm of a%x 2 AM, 
OD = p 2 = one-half the arm of a$y 2 AM, 
Oe = p = one-half the arm of the resultant couple. 

Let angle DOe — = GEK: we shall then have 
p = Oc = Of + fe = p 2 cos + p x sin ; 



COUPLES IN THE SAME OR PARALLEL PLANES. $J 

but cosO = f, sin0 = g, 

'" * R R 

:. Rp = F,p, + /^ a . 

Hence the moment of the resultant couple is equal to the sum 
of the moments of the separate couples, or 

R(2p) = a^r^AM= a%x 2 AM+ a%y 2 AM. 

Hence : To impart to a body a rotation about an axis passing 
through its centre of gravity requires the action of a statical 
couple, and conversely a statical couple so applied will cause such 
a rotation as that described. 

Hence we may generalize all our propositions in regard to 
the effect of statical couples and we may conclude that — 

In order that two couples may have the same effect, it is 
necessary — 

i°. That they be in the same or parallel planes. 

2°. That they have the same moment. 

3°. That they tejtd to cause rotation in the same direction 
(i.e., both right-handed or both left-handed when looked at from 
the same side). 

It also follows, that, for a given statical couple, we may sub- 
stitute another having the magnitudes of its forces different, 
provided only the moment of the couple remains the same. 

§ 57. Composition of Couples in the Same or Parallel 
Planes. — If the forces of the couples are not 
the same, reduce them to equivalent couples 
having the same force, transfer them to the 
same plane, and turn them so that their arms 
shall lie in the same straight line, as in Fig-. 
34; the first couple consisting of the force F FlG ' 34 ' 

at A and — F at B, and the second of F at B and —F at C. 



53 



APPLIED MECHANICS. 



The two equal and opposite forces counterbalance each other, 
and we have left a couple with force F and arm 

AC = AB + BC 

/. Resultant moment = F.AC = F(AB) + F{BC). 

Hence : The moment of the couple which is the resultant of 
two or more couples in the same or parallel planes is equal to 
the algebraic sum of the moments of the component couples. 

EXAMPLES. 

i. Convert a couple whose force is 5 and arm 6 to an equivalent 
couple whose arm is 3. Find the resultant of this and another couple 
in the same plane and sense whose force is 7 and arm 8 ; also find the 
force of the resultant couple when the arm is taken as 5. 



Solution. 

Moment of first couple = 5 x 6 = 30 

When arm is 3, force — ^ =10 

Moment of second couple = 7 x 8 = 56 
Moment of resultant couple = 30 -+- 56 = 86 
When arm is 5, force = - 8 ^ = 17J 

Given the following couples in one plane : — 



Force. 


Arm. 


Force 


12 


17 <| 




f 5 


3 


8 






5 


7 


Convert to equivalent 


8 


6 


9 


couples having the < 


6 


12 


12 


following : — 




10 
14 


9 

6 J 




4 

I 



Arm. 



20 

The first and the last three are right-handed ; the second, third, and 
fourth are left-handed. Find the moment of the resultant couple, and 
also its force when it has an arm 11. 






COUPLES IN PLANES INCLINED TO EACH OTHER. 59 

§58. Representation of a Couple by a Line From the 

preceding we see that the effect of a couple remains the same 
as long as — 

i°. Its moment does not change. 

2°. The direction of its axis (i.e., of the line drawn perpen- 
dicular to the plane of the couple) does not change. 

3°. The direction in which it tends to piake the body turn 
{right-handed or left-handed) remains the same. 

Hence a couple may be represented by drawing a line in 
the direction of its axis (perpendicular to its plane), and laying 
off on this line a distance containing as many units of length 
as there are units of moment in the couple, and indicating by a 
dot, an arrow-head, or some other means, in what direction one 
must look along the line in order that the rotation may appear 
right-handed. 

This line is called the Moment Axis of the couple. 

§ 59. Composition of Couples situated in Planes inclined 
to Each Other. — Suppose we have two couples situated 
neither in the same plane nor in parallel planes, and that we 
wish to find their resultant couple. We may proceed as fol- 
lows : Substitute for them equivalent couples with equal arms, 
then transfer them in their own plane respectively to such posi- 
tions that their arms shall _ Rt 
coincide, and lie in the ff\ 
line of intersection of the /'If 
two planes. / //f b 

This having been done, 'V. §5^T ^ Ss v 

let 00, (Fig. 35) be the °' /7>?\>^\ 

common arm, F and — F / 1 / ' ^^V^J^" 

the forces of one couple, F \ // ix 

F, and — F x those of the Y 

other. The forces F and fig. 35. 

F s have for their resultant R, and — F and — F, have —R T . 

Moreover, we may readily show that R and — R, are equal and 



60 APPLIED MECHANICS. 

parallel, both being perpendicular to 00,. The resultant of 
the two couples is, therefore, a couple whose arm is 00, and 
force F, the diagonal of the parallelogram on F and F„ so that 



R = sjF 2 + F 2 + 2i^ x cos 0, 

where is the angle between the planes of the couples. Now, 
if we draw from O the line Oa perpendicular to 00, and to F, 
and hence perpendicular to the plane of the first couple, and if 
we draw in the same manner Ob perpendicular to the plane of 
the second couple, so that there shall be in Oa as many units' 
of length as there are units of moment in the first couple, and 
in Ob as many units of length as there are units of moment in 
the second couple, we shall have — 

i°. The lines Oa and Ob are the moment axes of the two 
given couples respectively. 

2°. The lines Oa and Ob lie in the same plane with F and 
F„ this plane being perpendicular to 00,. 

3°. We have the proportion 

Oa: Ob = F. OO, : F,.00,= F: F,. 

4°. If on Oa and Ob as sides we construct a parallelogram, 
it will be similar to the parallelogram on F and F t . We shall 
have the proportion 

Oc : R = Oa: F = Ob : F, ; 

and since the sides of the two parallelograms are respectively 
perpendicular to each other, the diagonals are perpendicular to 
each other ; and since we have also 

Oc = R ' 0a and Oa = F. 00, ,\ Oc = R • OO t , 
F 

it follows that Oc is perpendicular to the plane of the resultant 
couple, and contains as many units of length as there are* units 
of moment in the moment of the resultant couple; in other 



COUPLE AND SINGLE FORCE IN THE SAME PLANE. 6 1 

words, Oc will represent the moment axis of the resultant 
couple, and we shall have 



Oc = sJOa 2 -f Ob 2 -j- 2 0a . Ob cos a Ob; 



or, if we let 



Oa = L, Ob = J/", 6V = (9, tf<9£ = 0, 
£ = \/Z 2 + M 2 + 2ZJ/ cos 0. 

This determines the moment of the resultant couple ; and, for 
the direction of its moment axis, we have 



sin aOc — — sin 6 
G 



and 



sin b Oc 



L . 

— sm 
G 



Hence we can compound and resolve couples just as we do 
forces, provided we represent the couples by their moment axes 



EXAMPLES. 

i. Given L = 43, M = 15, = 65 ° ; find resultant couple. 

2. Given L = 40, JZ = 30, 6> = 30 ; find resultant couple. 

3. Given L = 1, J/= 5, = .45° ; find resultant couple. 

§ 60. Resultant of a Couple and a Single Force in the 
Same Plane Let M (Fig. 36 or 37) be the moment of the 




r. 



-F 




J 



vF 



Fig. 36. Fig. 37. 

given couple, and let OF = F be the single force. For the 
given couple substitute an equivalent couple, one of whose 
forces is — F at O, equal and directly opposed to the single 



62 



APPLIED MECHANICS. 



force F, these two counterbalancing each other, and leaving 
only the other force of the couple, which is equal and parallel 
to the original single force F, and acts along a line whose 



distance from O is OA = — . 

F 



Hence 



The resultant of a single force and a couple in the same plane 
is a force equal and parallel to the origifial fo?re, having its 
line of direction at a perpendicular distance from the original 
force equal to the moment of the couple divided by the force. 

Fig. 36 shows the case when the couple is right-handed, and 
Fig. 37 when it is left-handed. 

§61. Composition of Parallel Forces in General. — In 
each case of composition of parallel forces (§§ 34, 35, and 36) 
it was stated that the method pursued was applicable to all 
cases except those where 

2F = o. 

We were obliged, at that time, to reserve this case, because we 
had not studied the action of a statical couple ; but now we will 
adopt a method for the composition of parallel forces which will 
apply in all cases. 



w 



When all the forces are in one plane. Assume, as we did 
in §35, the axis OY to be parallel to 
the forces ; assume the forces and the 
co-ordinates of their lines of direction, 
as shown in the figure (Fig. 38). Now 
place at the origin O, along OY, two 
equal and opposite forces, each equal to 
F l ; then these three forces, viz., F l at 
D, OA, and OB, produce the same effect 
as F t at D alone ; but F x at D and OB 
form a couple (left-handed in the figure) 
FlG38 - whose moment is —F.x^ Hence the 

force F x is equivalent to — 



Y 










i 


. 






. * 


4 


4 
1 


> 






< 


F, 


F 4 






C 






1 


D 




• 











COMPOSITION OF PARALLEL FORCES. 



63 



i°. An equal and parallel force at the origin, and 

2°. A statical couple whose moment is —F x x x . 

Likewise the force F 2 is equivalent to (i°) an equal and par- 
allel force at the origin, and (2 ) a couple whose moment is 
— /yr 2 , etc. 

Hence we shall have, if we proceed in the same way with 
all the forces, for resultant of the entire system a single force 

R = 2F along OY, 

and a single resultant couple 

M= -%Fx. 

(Observe that downward forces and left-handed couples are 
to be accounted negative.) 

Now, there may arise two cases. 
i°. When 2F= o, and 
2°. When ZFXo. 

Case I. When %F = o, the resultant force along OY van- 
ishes, and the resultant of the entire system is 
a statical couple whose moment is 



M = -%Fx. 

Case II. When %F > < o, we can reduce 
the resultant to a single force. 

Let (Fig. 39) OB represent the resultant 
force along OY, R = 2,F. With this, compound 
the couple whose moment is M = —%Fx, and 
we obtain as resultant (§ 60) a single force 

R = %F, 



ba 



c y 



Fig. 39. 



whose line of action is at a perpendicular distance from OY 
equal to 



AO = X r 



^F 



6 4 



APPLIED MECHANICS. 





z 








i 


F 


4 


« 




O 








^V. F 


\/ 


B 








A 




> 


f 









(/;) WJien the forces are not confined to 07ie plane. Assume, 
as before (Fig. 40), OZ parallel to the forces, and let F acting 

through A be one of the given 
forces, the co-ordinates of A be- 
ing x and y. Place at O two equal 
and opposite forces, each equal to 
F, and also at B two equal and 
opposite forces, each equal to F. 
These five forces produce the 
same effect as F alone at A, and 
they may be considered to con- 
sist of — 
i°. A single force F at the origin. 
2°. A couple whose forces are F at B and 
whose moment is — Fx acting in the y plane. 

3 . A couple whose forces are F at A and — F at B, and 
whose moment is Fy acting in the x plane. Treating each of 
the forces in the same way, we shall have, in place of the entire 
system of parallel forces, the following forces and couples : — 
i°. A single force R — %F along OZ. 
2°. A couple My — —2Fx in the y plane. 
3 . A couple M x — +2Fy in the x plane. 
Now, there may be 



Fig. 40. 



■F at O, and 




two cases : — 

i°. When %F >< o. 
2°. When $F = o. 



Case I. When %F> 
< o, we can reduce to a 
single resultant force 
having a fixed line of 
direction. Lay off (Fig. 
41) along OZ,OH=\%F. 
Combining this with the first of the above-stated couples, we 



COMPOSITION OF PARALLEI FORCES. 



65 



obtain a force R = %F at A, where OA 



%Fx 
^F 



== ;r r . Then 



combine with this resultant force R = ^F at A, the second 
couple, and we shall have as single resultant of the entire 
system a single force 

R = %F 
acting through B, where 

—if. 

Hence the resultant is a force whose magnitude is 

R = 2R 



the co-ordinates of its line of direction being: 



%Fx 



%Fy 



Case II. When ^F= o, there is no single resultant force ; 
but the system reduces to two couples, one in the x plane and 
one in the y plane, and these two can be reduced to one single 
resultant couple. (Observe that couples are to be accounted 
positive when, on being looked at by the observer from the posi- 
tive part of the axis towards 
the origin, they are right- 
handed ; otherwise they are 
negative.) 

The moment axis of the 
couple in the x plane will 
be laid off on the axis OX 
from the origin towards the 
positive side if the moment 
is positive, or towards the 

negative side if it is nega- Fig. 42. 

tive, and likewise for the couple in the y plane. 




66 



APPLIED MECHANICS. 



Hence lay off (Fig. 42) OB = M x , OA = ^, and by 
completing the rectangle we shall have OD as the moment 
axis of the resultant couple; hence the resultant couple lies 
in a plane perpendicular to OD, and its moment bears to 
OD the same ratio as M x bears to OB. 

Hence we may write 



OD = M r = <JM X > + M/, 

M r 



cos D OX 



cos 6. 



If My had been negative, we should have OE as the moment 
axis of the resultant couple. 



EXAMPLES. 



F. x. y. F. x. y. 

1. 5 4 3 2 - 5 -4 3 

3 2 1 —2 2—1 

13 5 -3 3 5 



Find the resultant in each example. 

§62. Resultant of any System of Forces acting at Dif- 
ferent Points of a Rigid 
Body, all situated in One 
Plane. — Let CF — F (Fig. 
43) be one of the given forces. 
Let all the forces be referred 
to a system of rectangular 
axes, as in the figure, and let 
a = angle made by F with 
OX, etc. Let the co-ordi- 



J 




M A 



Fig. 43. 



nates of the point of application of F be AO = x> BO = y. 



SYSTEM OF FORCES ACTING ON RIGID BODY. 6 J 

We first decompose CF = F into two components, parallel 
respectively to OX and OY. These components are 

CD = Fcosa, CE — Fsina. 

Apply at O in the line OY two equal and opposite forces, each 
equal to .Fsina, and at O in the line OX two equal and opposite 
forces, each equal to Fcosa. Since these four are mutually 
balanced, they do not alter the effect of the single force ; and 
hence we have, in place of Fat C, the six forces CD, OM, OK, 
CE, ON, OG. Of these six, CE and OG form a couple whose 
moment is 

— (Fsin a)x = —Ex sin a, 

CD and OK form a couple whose moment is 
(Fcosa)jp = Fy cos a. 

These two couples, being in the same plane, give as result- 
ant moment their algebraic sum, or 

E(y cos a — x sin a) . 

We have, therefore, instead of the single force at C, the follow- 
ing:— 

i°. OM = Fcos a along OX. 

2°. ON— Fsina along O Y. 

3 . The couple M == F(y cos a — x sin a) in the given plane. 

Decompose in the same way each of the given forces; and 
we have, on uniting the components along OX, those along OY, 
and the statical couples respectively, the following: — 

i°. A resultant force along OX, R x = SFcos a. 

2°. A resultant force along OY, R y = SFsina. 

3 . A resultant couple in the plane, whose moment is 

M = %F{y cos a — x sin a) . 



68 



APPLIED MECHANICS. 



This entire system, on compounding the two forces at O, 
reduces to 



i°. R = S/R x 2 + Rf = \I(%F cos a) 2 + (XFsina)*; 
making with 6Uf an angle a r , where 

2^ cos a 



cos a r = 



R 



2°. A resultant couple in the same plane, whose moment is 
M = %F{ y cos a — x sin a) . 

Now compound this resultant force and couple, and we have, 

for final resultant, a single 
force equal and parallel to 
R, and acting along a line 
whose perpendicular dis- 
tance from O is equal to 

M 
R' 



Fig. 44. 

Suppose (Fig. 44) the force 

OB = IF cos a, 
OA = XFsina, 



The equation of this line 
may be found as follows : 



OR = s/($Fcosa) 2 + (2F sin a) 2 ;' 

and let us suppose the resultant couple to be right-handed, and 

let 

OM=^: 
R 

then will the line ME parallel to OR be the line of direction 
of the single resultant force. 



CONDITIONS OF EQUILIBRIUM. 69 

Assuming the force R to act at any point C(x r ,y T ) of this 
line, if we decompose it in the same way as we did the single 
forces previously, we obtain — 

i°. The force R cos a r — %Fcos a along OX. 

2°. The force R sin<x r — S^sina along OY. 

3 . The couple R(y r cos a r — x r sin a r ). 

Hence we must have 

R(y r cos a r — x r sin a r ) = ^F(y cos a — x sin a) = M. 
Hence for the equation of the line of direction we have 

y r cos a r — ^ r sin cv = — . (1) 

-a 

Another form for the same equation is 

>(XFcosa) - x r (%F sin a) = J/". (2) 

§ 63. Conditions of Equilibrium. — If such a set of forces 
be in equilibrium, there must evidently be no tendency to trans- 
lation and none to rotation. Hence we must have 

R = o and M = o. 

Hence the conditions of equilibrium for any system of forces 
in a plane are three ; viz., — 

%F cos a = o, XFsina = o, *ZF{y cos a — ^sina) = o. 

Another and a very convenient way to state the conditions of 
equilibrium for this case is as follows : — 

If the forces be resolved into components along two directions 
at right angles to each other, then the algebraic sum of the com- 
ponents along each of these directions must be zero, and the 
algebraic sum of the moments of the forces about any axis per- 
pendicular to the plane of the forces must equal zero. 



7° 



APPLIED MECHANICS. 



EXAMPLES. 





F. 


#. 


y> 


a. 


r 


5 


3 


2 


3i' 


i. Given < 


IO 


i 


3 


40' 


{ 


-7 


4 


2 


54' 




F. 


*. 


7- 


a. 


f 


12 


27 


3 


•5 


2. Given < 


4 


J 3 


-5 


30 


[ 


8 


-5 


-4 


45 



Find the resultant, and 
the equation of its 
line of direction. 



I Find the resultant, and 
}■ the equation of its 
J line of direction. 



§ 64. Resultant of any System of Forces not confined 

to One Plane. — Suppose we 
have a number of forces applied 
at different points of a rigid 
body, and acting in different 
directions, of which we wish to 
find the resultant. Refer them 
all to a system of three rect- 
angular axes, OX, OY, OZ 
(Fig. 45). Let PR — F be 
one of the given forces. Re- 
solve it into three components, 
PK, PH, and PG, parallel 
Let 




Fig. 45- 

respectively to the three axes. 



RPK = a, RPH = p, RPG = y. 

Let OA = x, OB = y, OC = z, be the co-ordinates of the 
point of application of the force F. Now introduce at B and 
also at O two forces, opposite in direction, and each equal to PK. 
We now have, instead of the force PK, the five forces PK, BM, 
BN, OS, and OT. The two forces PK and BN form a couple 
in the y plane, whose axis is a line parallel to the axis OY, and 
whose moment is (PK)(EB) = (Fcos a )z = Fzcosa. The 



FORCES NOT CONFINED TO ONE PLANE. 



71 



forces BM and T form a couple in the z plane, whose moment 

is 

{BM)(OB) = -Fy cos a. 

Now do the same for the other forces PH and PG, and we shall 
finally have, instead of the force PR, three forces, 

F cos a, F cos /3, F cos y, 

acting at O in the directions OX, OY, and (9^ respectively; 
together with six couples, two of which are in the x plane, two 
in the y plane, and two in the z plane. 

They thus form three couples, whose moments are as fol- 
lows : — 

Around OX, F(y cos y — z cos ft) ; 

Around OY, F(z cos a — x cos 7); 

Around OZ, F(x cos ft —y cos a) . 

Treat each of the given forces in the same way, and we shall 
have, in place of all the forces of the system, three forces, 

2F cos a along OX, 
IF cos/? along OY, 
X^cosy along OZ; 

and three couples, whose moments are as follows : — 

Around OX, M x = 2^0' cosy - scos/3); 
Around OY, M y = %F(z cos a — *cosy); 
Around OZ, M z = %F(xcosfi —ycosa). 

The three forces give a resultant at O equal to 



R = s/ (IF cos a) 2 +■ (2,F cos /3) 2 + (2,F cosy) 2 , 



COS a r = 



IFcos 



cos/3, 



ZFcosft 
R ' 



cos y r 



2^ cosy 
R ' 



(1) 
(2) 



72 



APPLIED MECHANICS. 



For the three couples we have as resultant 



^ = sjM x 2 + Mf + MS, 



. M x 

COS A = — -, COSu 



My 

M' 



COS V = 



M } 



(3) 
(4) 



A, /x, and v being the angles made by the moment axis of the 
resultant couple with OX, O Y, and OZ respectively. 

T hits far we have reduced the whole system to a single result- 
ant force at the origin, and a couple. Sometimes we can reduce 

the system still farther, 
and sometimes not. The 
following investigation will 
show when we can do so. 
Let (Fig. 46) OP = R be 
the resultant force, and 
OC =M the moment axis 
of the resultant couple. 
Denote the angle between 
them by $ (a quantity thus 
far undetermined). Pro- 
ject OP — R on OC. Its 
projection will be OD = RcosO; then project, in its stead, the 
broken line OABP on OC. By the principles of projections, 
the projection of this broken line will equal OD. 

Now OA, AB, and BP are the co-ordinates of P, and make 
with c^ the same angle as the axes OX, OY, and OZ; i.e., 
X, fi, and v respectively : hence the length of the projection is 

(X4cosA -f AB cos \x + BP 'cos v. 

But 

OA — R cos a r , AB = R cos (B r , BP = R cos y r . 
Hence 

R cos 6 = R cos a r cos \ + R cos /? r cos /x + R cos y r cos V 

COS0 = COS (V COS A + COS /3 r COS {X + cos y r cos v. (5) 




Fig. 46. 



CONDITIONS OF EQUILIBRIUM. 73 

This enables us to find the angle between the resultant force 
and the moment axis of the resultant couple. 

The following cases may arise : — 

i°. When cos = o, or 6 — 90 , the force lies in the plane 
of the couple, and we can reduce to a single force acting at a 

distance from O equal to — , and parallel to R at O. 

2°. When cos = 1, or = o, the moment axis of the 
couple coincides in direction with the force: hence the plane 
of the couple is perpendicular to the force, and no farther 
reduction is possible. 

3 . When 6 is neither o° nor 90 , we can resolve the couple 
M into two component couples, one of which, Mcos 0, acts in a 
plane perpendicular to the direction of R, and the other, M sin 0, 
acts in a plane containing R. . The latter, on being combined 
with the force R at the origin, gives an equal and parallel force 
whose line of action is at a distance from that of R at O, equal 
to 

MsmO 
R ' 

4 . When M = o, the resultant is a single force at O. 

5°. When R = o, the resultant is a couple. 

§65. Conditions of Equilibrium. — To produce equilibrium, 
we must have no tendency to translation and none to rotation. 
Hence we must have 

R = o and M = o. 

Hence we have, in general, six conditions of equilibrium ; viz., — 

SZ^cosa = o, ^Fcos/S = o, 2LFcosy = o. 
M x = o, My = o, M z = o. 



74 APPLIED MECHANICS. 



EXAMPLES. 

i. Prove that, whenever three forces balance each other, they must 
lie in one plane. 

2. Show how to resolve a given force into two whose sum is given. 

3. A straight rod of uniform section and material is suspended by two 
strings attached to its ends, the strings being of given length, and attached 
to the same fixed point : find the position of equilibrium of the rod. 

4. Two spheres are supported by strings attached to a given point, 
and rest against each other : find the tensions of the strings. 

5 . A straight rod of uniform section and material has its ends resting 
against two inclined planes at right angles to each other, the vertical 
plane which passes through the rod being at right angles to the line of 
intersection of the two planes : find the position of equilibrium of the 
rod, and the pressure on each plane, disregarding friction. 

6. A certain body weighs 8 lbs", when placed in one pan of a false 
balance of equal arms, and 10 lbs. in the other : find the true weight of 
the body. 

7. The points of attachment of the three legs of a three-legged table 
are the vertices of an isosceles right-angled triangle ; a weight of 100 lbs. 
is supported at the middle of a line joining the vertex of one of the acute 
angles with the middle of the opposite side': find the pressure upon 
each leg. 

8. A heavy body rests upon an inclined plane without friction : find 
the horizontal force necessary to apply, to prevent it from falling. 

9. A rectangular picture is supported by a string passing over a 
smooth peg, the string being attached in the usual way at the sides, but 
one-fourth the distance from the top : find how many, and what are the 
positions of equilibrium, assuming the absence of friction. 

10. Two equal and weightless rods are jointed together, and form a 
right angle ; they move freely about their common point : find the 
ratio of the weights that must be suspended from their extremities, that 
one of them may be inclined to the horizon at sixty degrees. 

11. A weight of 100 lbs. is suspended by two flexible strings, one 
of which is horizontal, and the other is inclined at an angle of thirty 
degrees to the vertical : find the tension in each string. 



D YNAMICS. — DEFINITIONS. 



75 



CHAPTER II. 
DYNAMICS. 

§ 66. Definitions. — Dynamics is that part of mechanics 
which discusses the forces acting, when motion is the result. 

Velocity, in the case of uniform motion, is the space passed 
over by the moving body in a unit of time ; so that, if s repre- 
sent the space passed over in time t, and v represent the velocity, 
then 

s 

v = -. 
t 

Velocity, in variable motion, is the limit of the ratio of the 
space (Aj) passed over in a short time (A/), to the time, as the 
latter approaches zero : hence 

ds 

V — — . 

dt 



Acceleration is the limit of the ratio of the velocity (Av) im- 
parted to the moving body in a short time (A/), to the time, as 
the time approaches zero. Hence, if a represent the accelera- 
tion, 

dv = a \dt) = d^s 
dt dt dt 2 ' 






y6 APPLIED MECHANICS. 

§ 67. Uniform Motion In this case the acceleration is 

zero, and the velocity is constant ; and we have the equation 

s = vt. 

§ 68. Uniformly Varying Motion. — In this case the ac- 
celeration is constant : hence a is a constant in the equation 

d 2 s 

— = a , 

dt 2 

and we obtain by one integration 

ds , . 
v = — = at -f- c, 
dt 

where c is an arbitrary constant : to determine it we observe, 
that, if v Q represent the value of v when / = o, we shall have 

v Q = o + c 

.-. c — v Q 

ds , . 

" v = — = at + v , 
dt 

and by another integration 

s = ±at 2 -\- v Q t, 

where s is the space passed over in time t ; the arbitrary con- 
stant vanishing, because, when / = o, s is also zero. 

§ 69. Measure of Force. — It has already been seen, that, 
when a body is either at rest or moving uniformly in a straight 
line, there are either no forces acting upon it, or else the forces 
acting upon it are balanced. If, on the other hand, the motion 
of the body is rectilinear, but not uniform, the only unbalanced 
force acting is in the direction of the motion, and equal in mag- 
nitude to the momentum imparted in a unit of time in the direc- 
tion of .the motion, or, in other words, to the limit of the ratio 
of the momentum imparted in a short time (A/), to the time, as 
the latter approaches zero. 



MECHANICAL WORK. — UNIT OF WORK. JJ 

Thus, if i 7 denote the force acting in the direction of the 
motion, m the mass, and a the acceleration, we shall have 

„ dv d 2 s (i) 

F = ma = m — — m — . v ' 

dt dt 2 

From (i) we derive 

mdv = Fdt; (2) 

and, if v Q be the velocity of the moving body at the time when 
/ = t Q , and Vj, its velocity when t — t v we shall have 



J mdv — j 
v J to 



h 
Fdt 



or 

m(v, — v ) = t Fdt; (3) 

or, in words, the momentum imparted to the body during the 
time / = (t 1 — / ) by the force F, will be found by integrating 
the quantity Fdt between the limits t l and t Q . 

§ 70. Mechanical Work. — Whenever a force is applied to 
a moving body, the force is either used in overcoming resist- 
ances (i.e., opposing forces, such as gravity or friction), and 
leaving the body free to continue its original motion undis- 
turbed, or else it has its effect in altering the velocity of the 
body. In either case, the work done by the force is the prod- 
uct of the force, by the space passed through by the body in 
the direction of the force. 

Unit of Work. — The unit of work is that work which is 
done when a unit of force acts through a unit of distance in 
the same direction as the force ; thus, if one pound and one 
foot are our units of force and length respectively, the unit of 
work will be one foot-pound. 

If a constant force act upon a moving body in the direction 
of its motion while the body moves through the space s, the 
work done by the force is 

Fs; 



7 8 APPLIED MECHANICS. 

and this, if the force is unresisted, is the energy, or capacity for 
performing work, which is imparted to the body upon which the 
force acts while it moves through the space s. 

Thus, if a io-pound weight fall freely through a height of 
5 feet, the energy imparted to it by the force of gravity during 
this fall is io X 5 = 50 foot-pounds, and it would be necessary 
to do upon it 5c foot-pounds of work in order to destroy the 
velocity acquired by it during its fall. If, on the other hand, 
the force is a variable, the amount of work done in passing 
over any finite space in its own direction will be found by in- 
tegrating, between the proper limits, the expression 

JFds. 

The power which a machine exerts is the work which it 
performs in a unit of time. 

The unit of power commonly employed is the horse-power, 
which in English units is equal to 33000 foot-pounds per 
minute, or 550 foot-pounds per second. 

§71. Energy. — The energy of a body is its capacity for 
performing work. 

Kinetic or Actual Energy is the energy which a body pos- 
sesses in virtue of its velocity ; in other words, it is the work 
necessary to be done upon the body in order to destroy its 
velocity. This is equal to the work which would have to be 
done to bring the body from a state of rest to the velocity with 
which it is moving. Assume a body whose mass is m, and sup- 
pose that its velocity has been changed from ^ O 'to v x . Then if 
F be the force acting in the direction of the motion, we shall 
have, from equation (2), § 69, that 

Fvdt = tnvdv; (1) 

but 

vdt = ds 

.*. Fds = mvdv. (2) 



AT WOOD'S MACHINE. ?g 

Hence, by integration, 

Jtnvdv = I Fds 

.'. \m{y* - vj) = /Fds; (3) 

but /Fds is the work that has been done on the body by the 
force, and the result of doing this work has been to increase 
its velocity from v Q to v x . It follows, that, in order to change 
the velocity from v Q to v„ the amount of work necessary to per- 
form upon the body is 

f»(* f » - v 2 ) = \—{v? - v Q 2 ). (4) 

g 

If v Q = o, this expression becomes 

imv*, or S (5) 



which is the expression for the kinetic energy of a body of mass 
m moving with a velocity v x . 

§ 72. Atwood's Machine. — A particular case of uniformly 
accelerated motion is to be found in Atwood's machine, in which 
a cord is passed over a pulley, and is loaded with unequal weights 
on the two sides. Were the weights equal, there would be no 
unbalanced force acting, and no motion would ensue ; but when 
they are unequal, we obtain as a result a uniformly accelerated 
motion (if we disregard the action of the pulley), because we 
have a constant force equal to the difference of the two weights 
acting on a mass whose weight is the sum of the two weights. 
Thus, if we have a 10-pound weight on one side and a 5-pound 
weight on the other, the unbalanced force acting is 

^=10 — 5 = 5 lbs. 



80 APPLIED MECHANICS. 



IO I ■ *" 

The mass moved is M = ■ — £ : hence the resulting ac- 

g 
celeration is 

a= -J--4 



(?) 



3 



§ 73. Normal and Tangential Components of the Forces 
acting on a Heavy Particle. — If a body be in motion, either 
in a straight or in a curved line, and if at a certain instant all 
forces cease acting on it, the body will continue to move at a 
uniform rate in a straight line tangent to its path at that point 
where the bodv was situated when the forces ceased acting:. 

If an unresisted force be applied in the direction of the 
body's motion, the motion will still take place in the same 
straight line ; but the velocity will vary as long as the force 
acts, and, from what we have seen, the equation 

F = w — (1) 

will hold. 

If an unresisted force act in a direction inclined to the 
body's motion, it will cause the body to change its speed, and 
also its course, and hence to move in a curved line. Indeed, 
if a force acting on a body which is in motion be resolved into 
two components, one of which is tangent to its path and the 
other normal, the tangential component will cause the body to 
change its speed, and the normal component will cause it to 
change the direction of its motion. 

The measure of the tangential component is, as we have 

seen, 

7T d 2 s 

F = m — ; 

dr- 

and we will proceed to find an expression for the normal com- 
ponent otherwise known as the Deviating Force. For this 



CENTRIFUGAL FORCE. 8 1 

purpose we may substitute, for a small portion of the curve, a 
portion of the circle of curvature ; hence we will proceed to 
find an expression for the centrifugal force of a body which 
moves uniformly with a velocity v in a circle whose radius is r. 

CENTRIFUGAL FORCE. 

Let AC (Fig. 47) be the space described in the time At. 

Then we have A B 

AC = vAt. s* 

The motion AC may be approximately consid- / 
ered as the result of a uniform motion I 

AB = vAt nearly, \/ 

and a uniformly accelerated motion fig. 47. 

BC = ia(At) 2 = s, 

where a = acceleration due to centrifugal force. But 

(AB) 2 = BC .BD, 
or 

(vAt)* = ±a{Aty(2r + s), 
where 

AO = OC = r 

/. v 2 = \a(zr -f s) approximately 
,\ a = approximately. 

For its true value, pass to the limit where s = o. 

Hence we have, for the acceleration due to the centrifugal 

force, the expression 

!? 
r 

Hence the centrifugal force is equal to 

„ mv^ _ Wv 2 , v 

r gr ' 



82 APPLIED MECHANICS. 



DEVIATING FORCE. 

If a body is moving in a curved path, whether circular or 
not, and the unbalanced force acting on it be resolved into tan- 
gential and normal components, the tangential component will 
be, as has already been seen, 

d 2 s 

m — : 

dt 2 

and the normal component will be 

mv 2 _ m/ds\ 2 
~ ~ ~r\dt) ' 

where r is the radius of curvature of the path at the point in 
question. 



RESULTANT FORCE. 

Hence it follows that the entire unbalanced force acting on 
the body will be 



or 






§ 74. Components along Three Rectangular Axes of the 

Velocities of, and of the Forces acting on, a Moving 

ds 
Body. — If we resolve the velocity — into three components 

dt 

along OX, OY, and OZ, we shall have, for these components 

respectively, 

d A ^, and A 

dt dt dt' 

this being evident from the fact that dx, dy, and dz are respec- 



COMPONENTS OF VELOCITIES AND FORCES. 83 

tively the projections of ds on the axes OX, OY, and OZ ; and, 
from the differential calculus, we have 



s -m +&+&■ 



On the other hand, 



dx dy , dz 

— , -*- t and — 
dt dt' dt 



are not only the components of the velocity — in the directions 

dt 

OX, OY, and OZ, but they are also the velocities of the body 

in these directions respectively. 

Now, the case of the accelerations is different; for, while 



d 2 x d 2 y j d 2 z 

, —£-. and 

dP dt 2 - dt 2 



are the accelerations in the directions OX, OY, and OZ respec- 
tively, they are not the components of the acceleration 

d^s 

dt 2 

along the three axes. 

That they are the former is evident from the fact that — , 

dt 

dv dz 

-f-, and — are the velocities in the directions of the axes, and 

at at 

d 2 x d 2 y d 2 z 

-r-, — , ~ are their differential co-efficients, and hence repre- 
sent the accelerations along the three axes. But if we consider 
the components of the force acting on the body, we shall have 



84 APPLIED MECHANICS. 

for its components along OX, OY, and OZ, if a, /3, and y are 
the angles made by F with the axes respectively, 

77 d 2 x z- /> d 2 y j? d 2 z 

/'cos a = m , FcosB — ??i-^-, I 1 cosy = ;;* , 

dt 2 ' H dt 2 ' Y dt 2 ' 



<■■' =nfJ t (?)'*(S)' <■> 

and we found (§ 73) for i 7 , the value 



W(£J+r.(!)- <•> 

Hence, equating these values of F, and simplifying, we shall 
have the equation 

Hence it is plain that — -, —=£, and —^ can only be the com- 
F ^ 2 ^ 2 ^ 2 

ponents of the actual acceleration 

d 2 s 

dt 2 

when the last term — (— ) vanishes, or when r = 00 , i.e., when 
r 2 \dt) 

the motion is rectilinear. 

Moreover, we have the two expressions (1) and (2) for the 
force acting upon a moving body. 

The truth of the proposition just proved may also be seen 
from the following considerations : — 

If a parallelopiped be constructed with the edges 

dx dy dz 
dt' dt' dt' 



CENTRIFUGAL FORCE OF A SOLID BODY. 85 

the diagonal will be the actual velocity 

ds_ 
dt* 

and will, of course, coincide in direction with its path. 

On the other hand, if a parallelopiped be constructed with 

the edges ' 

d 2 x d 2 y tPz 
dt 2 ' dt 2 ' dt 2 ' 

its diagonal must coincide in direction with the force 
F 



= 4&h¥$i' 



and can coincide in direction with the path, and hence with the 
actual acceleration 

d 2 s 

dt 2 ' 

only when the force is tangential to the path, and hence when 
the motion is rectilinear. 

§75. Centrifugal Force of a Solid Body. — When a solid 
body revolves in a circle, the resultant centrifugal force of the 
entire body acts in the direction of the perpendicular let fall 
from the centre of gravity of the body on the axis of rotation, 
and its magnitude is the same as if its entire weight were con- 
centrated at its centre of gravity. 

Proof. — Let (Fig. 48) the angular velocity = a, and the total 
weight = IV. Assume the axis of rotation perpendicular to 
the plane of the paper and passing through 
O ; assume, as axis of x, the perpendicular 
dropped from the centre of gravity upon 
the axis of rotation. The co-ordinates of 
the centre of gravity will then be (x OJ j/ ), 
and y Q will be equal to zero. 

If, now, P be any particle of weight w, 
where r = perpendicular distance from P on axis of rotation, 




86 APPLIED MECHANICS. 

and x == OA, y = ^P, we shall have for the centrifugal force 
of the particle at P 

w 2 
— a 2 r; 

g 

but if we resolve this into two components, parallel respectively 
to OX and O Y, we shall have for these components 

/W , \X a 2 , /W , \ V a 2 

[-a 2 r\- = —wx and {—a 2 ry- = -wy, 
\g / r g \g / r g 

and, for the resultant for the entire body we shall have, parallel 
to OX, 



a ~_, a 



F x = —%wx = —JVx , (i) 

g i 

and 

F y = --%wy = -Wy = o. (2) 

g g 

Hence the centrifugal force of the entire body is 

F= a -Wx - (3) 

g 

and if we let v Q = ax Q = linear velocity of the centre of gravity, 

we have 

Wv 2 



F = 



g*o 



which is the same as though the entire weight of the body 
were concentrated at its centre of gravity. 



EXAMPLES. 

i. A 10-pound weight is fastened by a rope 5 feet long to the 
centre, around which it revolves at the rate of 200 turns per minute ; 
find the pull on the cord. 

2. A locomotive weighing 50000 lbs., whose driving-wheels weigh 
7000 lbs., is running at 60 miles per hour, the diameter of the drivers 



UNIFORMLY VARYING RECTILINEAR MOTION 8y 

being 6 feet, and the distance from the centre of the wheel to the centre 
of gravity of the same being 2 inches (the drivers not being properly 
balanced) ; find the pressure of the locomotive on the track {a) when 
the centre of gravity is directly below the centre of the wheel, and {I?) 
when it is directly above. 

3. Assume the same conditions, except that the distance between 
centre of the wheel and its centre of gravity is 5 inches instead of 2. 

§76. Uniformly Varying Rectilinear Motion. — We have 
already found for this case (§ 68) the equations 



d 2 s 
dt 2 


= 


a 


= a constant, 


ds 
7/ 


= 


V 


= v + at, 


s 


= 


Vc 


t + \aP 1 



and we may write for the force acting, which is, of course, coin- 
cident in direction with the motion, 

r = m — = ma — a constant. 
dt 2 

§ yj. Motion of a Body acted on by the Force of Gravity 
only. — A useful special case of uniformly varying motion is 
that of a body moving under the action of gravity only. 

The downward acceleration due to gravity is represented by 
g feet per second, the value of g varying at different points on 
the surface of the earth according: to the following law : — 



= 6i C 1 — 0.00284 cos 2/\)/'i — —j feet per second, 



g t = 3 2. 1 695 feet, 

A = latitude of the place, 

h = its elevation above mean sea-level in feet, 

R == 20900000 feet. 



88 APPLIED MECHANICS. 

If, now, we represent by h the height fallen through by a 
descending body in time t, we shall have the equations, 

v = v -f gt, 
h = v Q t + \gt\ 

where v Q is the initial downward velocity. 

If, on the other hand, we represent by v Q the initial upward 
velocity, and by h the height to which the body will rise in 
time t under the action of gravity only, we must write the equa- 
tions 

v = v - gt, 

h = V Q t — \gt 2 . 

When v Q = o, the first set of equations gives 

v == gt, 
h = W\ 

which express the law of motion of a body starting from rest 
and subject to the action of gravity only. 

Eliminate t between these equations, and we shall have 

v 7 - = 2gh .'. v = \2gh, 
or 

*/* 

h — — : 

h is called the Jieight due to the velocity v, and represents the 
height through which a falling body must drop to acquire the 
velocity v ; and 

v = ^2gh 



UNRESISTED PROJECTILE. 89 

is the velocity which a falling body will acquire in falling 
through the height h. Thus, if a body fall through a height of 
50 feet, it will, by that fall, acquire a velocity of about 



^2(32^) (50) = V/3216. 66 = 56.7 feet per second. 

Again : if a body has a velocity of 40 feet per second, we shall 
have 

, v 1 1600 _ r . 

fi = — = = 24.8 feet ; 

*g 64.3 

and we say that the body has a velocity due to the height 24.8 
feet, i. e., a velocity which it would acquire by falling through a 
height of 24.8 feet. 



EXAMPLES. 

1. A stone is dropped down a precipice, and is heard to strike the 
bottom in 4 seconds after it started : how high is the precipice? 

2. How long will a stone, dropped down a precipice 500 feet high, 
take to reach the bottom? 

3. What will be its velocity just before striking the ground? 

4. A body is thrown, vertically upwards with a velocity of 100 feet 
per second ; to what height will it rise ? 

5. A body is thrown vertically upwards, and rises to a height of 50 
feet. With what velocity was it thrown, and how long was it in its 
ascent? 

6. What will be its velocity in its ascent at a point 15 feet above 
the point from which it started, and what at the same point in its 
descent ? 

§ 78. Unresisted Projectile. — In the case of an unresisted 
projectile, we have a body on which is impressed a uniform 



9 o 



APPLIED MECHANICS. 



motion in a certain direction (the direction of its initial motion), 
and which is acted on by the force of gravity only. 

Let OPC be 
the path (Fig. 49), 
OA the initial di- 
rection, and v Q the 
initial velocity, and 
the angle AOX = 

Then we shall 
have, for the hori- 
zontal and vertical 

components of the unbalanced force acting, when the projectile 

is at P (co-ordinates x and y), 




Fig. 49. 



m 



dt 2 



d 2 y 
o along OX, and m — = 
b dt 2 



mg — —W along OY. 



Hence 



d 2 x 
dt~ 2 



o, 



(1) 



d 2 y 
dP 



-g- 



(2) 



Integrating, and observing, that, when t = o, the horizontal 
and the vertical velocities were respectively x> cos 6 and ^ sin 0, 
we have 

X = v cos 0, (3) 



dt 
dt 



v Q sin 



-** 



(4) 



These equations could be derived directly by observing that 
the horizontal component of the initial velocity is v Q cos 0, and 
that this remains constant, as there is no unbalanced force act- 
ing in this direction, also that ^ sin^ is the initial vertical 
velocity ; and, since the body is acted on by gravity only, this 
velocity will in time t be decreased by gt. 



UNRESISTED PROJECTILE. 9 1 

Integrating equations (3) and (4), and observing that for 
/ z= o, x and y are both zero, we obtain 

x — v cos 6.t, (5) 

y = v sin B.t - \gt 2 . (6) 

Eliminate t, and we have 

y = xianO — — ■ (7) 

2^ o 2 C0S 2 6> 

as the equation of the path, which is consequently a parabola. 

Equations (1), (2), (3), (4), (5), (6), and (7) enable us to solve 
any problem with reference to an unresisted projectile. 

Equation (7) may be written 

( — Z '° 2 S " 12 \ == £_ - ( X — V ° 2 sin ^ C0S ^ V (Q\ 

V 2^ / 2^ 2 COS 2 $ \ g ) ^ ' 

which gives for the co-ordinates of the vertex 

v 2 sin 2 6 v 2 sin 6 cos 

yi = , x x = . 

2g g 

EXAMPLES. 

i. An unresisted projectile starts with a velocity of 100 feet per 
second at an upward angle of 30 to the horizon ; what will be its velocity 
when it has reached a point situated at a horizontal distance of 1000 feet 
from its starting-point, and how long will be required for it to reach 
that point? 

Solution. 

v = 100, = 30 , v Q cos 6 = 86.6, v sin 6 = 50, 

g = 32.16. 
Equation (5) gives us 

1000 = 86.6 / 

. 1000 , 

/. / = — - = 11.55 seconds. 

00.0 



9 2 



APPLIED MECHANICS. 



z/ sin<9 - gt = 50 — 371.5 = -321.5. 



v = V(86.6) 2 -f (321.5) 2 = ^75°° + 103362 = 333- 

Hence the point in question will be reached in ii|- seconds after start- 
ing, and the velocity will then be ^^ feet per second. 

2. An unresisted projectile is thrown upwards from the surface of 
the earth at angle of 39 to the horizontal : find the time when it will 
reach the earth, and the velocity it will have acquired when it reaches 
the earth, the velocity of throwing being 30 feet per second. 

3. A 10-pound weight is dropped from the window of a car when 
travelling over a bridge at a speed of 25 miles an hour. How long will 
it take to reach the ground 100 feet below the window, and what will be 
the kinetic energy when it reaches the ground ? 

4. With what horizontal velocity, and in what direction, must it be 
thrown, in order that it may strike the ground 50 feet forward of the 
point of starting? 

5. Suppose the same 10-pound weight to be thrown vertically up- 
wards from the car window with a velocity of 100 feet a minute, how 
long will it take to reach the ground, and at what point will it strike the 
ground? 

§ 79. Motion of a Body on an Inclined Plane without 

Friction. — If a body move on 
an inclined plane along the line 
of steepest descent, subject to 
the action of gravity only, and 
if we resolve the force acting 
on it (i.e., its' weight) into two 
components, along and perpen- 
dicular to the plane respec- 
tively, the latter component 
will be entirely balanced by 
FlG -5o. the resistance of the plane, 

and the former will be the only unbalanced force acting on 

the body. 




MOTION OF A BODY ON AN INCLINED PLANE. 93 

Suppose a body whose weight is represented (Fig. 50) by 
HF = W to move along the inclined path AB under the action 
of gravity only. Let be the inclination of AB to the horizon. 
Resolve Winto two components, 

HD = Wsm 0, and HE = W cos 6, 

respectively parallel and perpendicular to the plane. The 
former is the only unbalanced force acting on the body, and 
will cause it to move down the plane with a uniformly accel- 
erated motion ; the acceleration being 

= £-sin0. (1) 



(?) 



If the body is either at rest or moving downwards at the 
beginning, it will move downwards ; whereas, if it is first mov- 
ing upwards, it will gradually lose velocity, and move upwards 
more slowly, until ultimately its upward velocity will be de- 
stroyed, and it will begin moving downwards. 

The equations for uniformly varying motion are entirely 
applicable to these cases. Thus, suppose that the body has an 
initial downward velocity v Q1 this velocity will, at the end of the 
time t, become 

v = -z- = z>o+ (gsm0)t (2) 

at 

/. s = vj + k sin . t 2 , (3) 

and, for the unbalanced force acting, we have 

F=m^ = —{g sin0) = WsmO. (4) 

at 2 ^ 



94 APPLIED MECHANICS. 

If, on the other hand, the body's initial velocity is upward, 
and we denote this upward velocity by v Q , we shall have the 
equations 

V = ~ = v ■.— (gsmO)t (5) 

s = V - A?sin0 . t 2 (6) 

/r = _#/ s in0. (7) 

Again, if the initial velocity is zero, equations (2) and (3) 
become 

v = ^ = (gsm6)t, (8) 

at 



From these we obtain, for this case, 



\/; 



g sin#' 
and, substituting this value of / in (8), we have 



(9) 



(10) 



v = SJ2g{s sin 6/), (11) 

or, if we let s sin = h — the vertical distance through which 
the body has fallen, we have 

v = ^2gh. (12) 

Hence, When a body, starting from rest,, falls, under the 
action of gravity only, through a height h, the velocity acquired 
is V^2gh, whether the path be vertical or inclined. 

EXAMPLES. 

i. A body moves from the top to the bottom of a plane inclined 
to the horizon at 30 , under the action of gravity only : find the time 
required for the descent, and the velocity at the foot of the plane. 



MOTION ALONG A CURVED LINE. 



95 



2. In the right-angled triangle shown in the figure (Fig. 51 
AB = 10 feet, angle BAC = 30 : find the time a 
body would require, if acted on by gravity only, to fall 
from rest through each of the sides respectively, AB 
being vertical. 

3. Given inclination of plane to the horizon = 0, 
length of plane == /.• compare the time of falling down 
the plane with the time of falling down the vertical. 

4. A 100-pound weight rests, without friction, on the 
plane of example 3. What horizontal force is required 
to keep it from sliding down the plane. 

5. Suppose 5 pounds horizontal force to be applied 
(a) so as to oppose the descent, (b) so as to aid the descent : 
each case how long it will take the weight to descend from the 
the bottom plane. 




§ 80. Motion along a Curved Line under the Action of 
Gravity only. — We shall consider two questions in this 
regard : (a) the velocity at any point of the curve (b) the time 
of descent through any part of the curve. 

(a) Velocity at any point. Let us suppose the body to have 

started from rest at A, and to have 
reached the point P in time /, 
where AB = x (Fig. 52). Then, 
since the curved line AP may be 
considered as the limit of a broken 
line running from A to P, and as 
it has already been seen that the 
velocity acquired by falling through 
a certain height depends only upon 

Fig. 52. . . . ' & _ r , J . . . 

the height, and not upon the incli- 
nation of the path, we shall have for a curved line also 




v = y2gAB — 
where v is the velocity at P. 



2gX, 



9 6 



APPLIED MECHANICS. 



(b) Time down a curve. Referring to the same figure, let t 
denote the time required to go from A to P, and At the time to 
go from P to P', where PP' = As, and BB' — Ax; then, as we 
have seen that the velocity at P is ^2gx, we shall have approx- 
imately for the space passed over in time At, the equation 



or, passing to the limit, 



This equation gives 



or 



As = V ' 2gxAt, 



ds _ i 

7/ ~ ^ 



dt 



ds 



ds 



vV 



J^2gX J 



\MiT- 



(0 



(2) 



2gX 



where, of course, the proper limits of integration must be 
used. 

If t denote the time from A to P. we have 



/: 



ds 



2SX 




Fig. 53. 



EXAMPLE. 

A body acted on by gravity only is constrained to 
move in the arc of a circle from A to C (Fig. 53), radius 
10 feet. Find the time of describing the arc (quadrant) 
and the velocity acquired by the body when it reaches- 
C. 



SIMPLE CIRCULAR PENDULUM. 



97 



§81. Simple Circular Pendulum. — To find the time occu- 
pied in a vibration of a simple circu- ^c 
lar pendulum, we take D (Fig. 54) as 
origin, and DC as axis of x, and the 
axis of y at right angles to DC. Let 
AC = /and BD = /z, we shall have 
for the time of a single oscillation 
from A to E 



/x = h 



<& 




Now, from the equation of the circle AFDB, 

y 2 = 2/x — x 2 , 



we have 



*/y _ I — x 
dx y 

ds I I 



^ y \J 2 i x 



/ax _ _2£_ r dx 

^(2/x - x 2 )[2g(k - *)] ^2g J o \jhx - x 2 slzl—x 
or 

"V^J s/hx - x 2 \ 2/) 



This can only be integrated approximately. 
Expanding ( 1 — — J 



we obtain 



1 +£+A£ + ctc v 

4/ 32 / 2 



(-3)" 

V^J V V 32 1* Nhx - x* 



98 APPLIED MECHANICS. 

The greatest value of x is h; and if h is so small that we may 
omit — , we shall have as our approximate result 

If, however, the value of h as compared with / is too large 

to render it sufficiently accurate to omit —, but so small that 

4/ 

we can safely omit the higher powers of -, we shall have 

t = i/-^versm 1 / - } 

V g{ h 4/J o sJh x -x 2 S o 

A II i . ~ x 2X , if/z . ~ x 2X . n :~1 ) h 

= V/-<versm 1 versm \hx — x 2 \\ 

V g \ h 4l[_2 _ h J j 

or 



a nearer approximation. 
The formula 

is the most used, and is more nearly correct, the smaller the 
value of h. 

EXAMPLES. 

i . Find the length of the simple circular pendulum which is to beat 
seconds at a place where g = 32J. 

Solution. 

-4 - '-sr-dg^— ■ 



SIMPLE CYCLOIDAL PENDULUM. 



99 



2. What is the time of vibration of a simple circular pendulum 5 
feet long? 



§ 82. Simple Cycloidal Pendulum, 
cycloid is 



The equation of the 



y = tfversin — \- (2ax — x 2 )*, 



ty _ * 2a ~ x 

dx V x 

ds^ _ /^fV 

dx \ x J 

Hence we shall have, for the time of a single oscillation, 

■ t= Mf - 



Shx 



or 



©M . _I 2x) k .fa 



This expression is independent of /z, so that the time of vibra- 
tion is the same whether the arc be large or small. 

A body can be made to vibrate in a cycloidal arc by suspend- 
ing it by a flexible string between two cycloidal cheeks. This 
is shown from the fact that 
the evolute of the cycloid is 
another cycloid (Fig. 55). 

To prove this, we have, 

from the equation of the 

cycloid, 

—1 x 
y = a versin — \- (2ax — x 2 )*, 



dy _ 1 2a — x ds _ 1 2a 
dx V x ' dx V x ' 
d 2 y _ 




dx 2 xl\]\ 



100 APPLIED MECHANICS. 

Hence the radius of curvature is 

k 3 



(I)' 



p = — — = (2)^(0)^20 — x 

— d 2 y 

dx 2 
or 

p = 2(2a)*^2a — x; 

and since we have for the evolute the relation 

ds' = dp, 

where ds' is the elementary arc of the evolute, 

x= x 

and, observing that when x = 2a p = o, we have 



.*. / = 2(2#)V2tf — x ; 

or, if we transform co-ordinates to B by putting x for 2a — x, 
we obtain 

/ = 2(2dX)*, 

/. / 2 = 8ax, 

which is the equation of another cycloid just like the first. 

The motion along a vertical cycloid may also be obtained by 
letting a body move along a groove in the form of a cycloid 
acted on by gravity alone ; and in this case the time of descent 
of the body to the lowest point is precisely the same at what- 
ever point of the curve the body is placed. 

§ 83. Effect of Grade on the Tractive Force of a Rail- 
way Train. — As a useful particular case of motion on an 
inclined plane, we have the case of a railroad train moving up 
or down a grade. It is necessary that a certain tractive force 



EFFECT OF GRADE ON TRACTIVE FORCE. 101 

be exerted in order to overcome the resistances, and keep 
the train moving at a uniform rate along a level track. If, 
on the other hand, the track is not on a level, and if we 
resolve the weight of the train into components at right angles 
to and along the plane of the track, we shall have in the latter 
component a force which must be added to the tractive force 
above referred to when we wish to know the tractive force re- 
quired to carry it up grade, and must be subtracted when we 
wish to know the tractive force required to carry it down grade. 
The result of this subtraction may give, if the grade is suffi- 
ciently steep and the speed sufficiently slow, a negative quan- 
tity ; and in that case we must apply the brakes, instead of 
using steam, unless we wish the speed of the train to increase. 

EXAMPLES. 

i. A railroad train weighing 60000 lbs., and running at 50 miles per 
hour, requires a tractive force of 6 1 8 lbs. on a level ; what is the tractive 
force necessary when it is to ascend a grade of 50 feet per mile? What 
when it is to descend? Also what is the amount of work per minute 
in each case ? 

Solution. 

The resolution of the weight will give (Fig. 50, § 79), for the com- 
ponent along the plane, 

( 6oooo )^b == 568.2 nearly. 
Hence 

Tractive force for a level = 618.0, 
Tractive force for ascent = 1186.2, 
Tractive force for descent = 49.8. 

To ascertain the work done per minute in each case, we have — 
(a) For a level track, 6l8 x 5 6 ° q x 528 ° = 2719200 foot-lbs. 

{b) Up grade, 2719200 + 6ooo ° * o 5 ° x 5 ° = 5219200 foot-lbs. 
(c) Down grade, 2719200 - 6ooo ° x ^° x 5 ° = 219200 foot-lbs. 



102 



APPLIED MECHANICS. 



2. Suppose the tractive force required for each 2000 lbs. of weight 
of train to be, on a level track, for velocities of — 

5.0 miles per hour, 10.0 20.0 30.0 40.0 50.0 60 

6.1 lbs., 6.6 8.3 11. 2 15.3 20.6 27; 
find the tractive force required to carry the train of example 1 — 

(a) Up an incline of 50 feet per mile at 30 miles per hour. 
{&) Down an incline of 50 feet per mile at 30 miles per hour. 

(c) Down an incline of 10 feet per mile at 20 miles per hour. 

(d) What must be the incline down which the train must run to 
require no tractive force at 40 miles per hour? 

3. If in the first example the tractive force remains 618 lbs. while 
the train is going down grade, what will be its velocity at the end of one 
minute, the grade being 10 feet per mile? 



§84. Harmonic Motion. — If we imagine a body to be 
moving in a circle at a uniform rate (Fig. 56), and a second 

body to oscillate back and forth in 

the diameter AB, both starting 

from B, and 

if when the 

first body is 

at C the other 

is directly un- 
der it at G, 

etc., then is 

the second 

body said to 
move in harmonic motion. 

A practical case of this kind of mo- 
tion is the motion of a slotted cross-head 
of an engine, as shown in the figure 
(Fig. 57); the crank moving at a uni- 
form rate. In the case of the ordinary 
crank, and connecting-rod connecting 
the drive-wheel shaft of a stationary engine with the piston-rod, 




Fig. 56. 




HARMONIC MOTION. 



we have in the motion of the piston only an approximation to 
harmonic motion. We will proceed to determine the law of the 
force acting upon, and the velocity of, a body which is con- 
strained to move in harmonic motion. Let the body itself and 
the corresponding revolving body be supposed to start from 
B (Fig. 56), the latter revolving in left-handed rotation with an 
angular velocity a, and let the time taken by the former in 
reaching G be t: then will the angle BOC = at; and we shall 
have, if s denote the space passed over by the body that moves 
with harmonic motion, 



or, if 



s = BG = OB — OC cos a/, 

r= OB = OC, 

s = r — r cos at, 



(0 



the velocity at the end of the time / will be 

ds 



V = — = arsmat, 
dt 



(2) 



and the acceleration at the end of time / will be 



J = = aV COS at. 

dt 2 



(3) 



Hence the force acting upon the body at that instant, in the 
direction of its motion, is 



d 2 s 
F = *»— = ?na 2 rcosat = ma 2 (0G). (4) 



The force, therefore, varies directly as the distance of the body 
from the centre of its path. It is zero when the body is at the 



104 APPLIED MECHANICS. 

centre of its path, and greatest when it is at the ends of its 

travel, as its value is then 

W 
ma?r = — a 2 r; 

g 

this being the same in amount as the centrifugal force of the 
revolving body, provided this latter have the same weight as the 
oscillating body. On the other hand, the velocity is greatest 

when at = - (i.e., at mid-stroke) ; and its value is then 

2 

V = ar, 

this being also the velocity of the crank-pin at mid-stroke. 

EXAMPLE. 

Given that the reciprocating parts of an engine weigh ioooo lbs., 
the length of crank being i foot, the crank making 60 revolutions per 
minute ; find the force required to make the cross-head follow the crank, 
(1) when the crank stands at 30 to the line of dead points, (2) when 
at 6o°, (3) when at the dead point. 

§85. Work under Oblique Force. — If the force act in 
any other direction than that of the motion, we must resolve it 
into two components, the component in the direction of the 
motion being the only one that does work. Thus if the force 
F is variable, and equals the angle it makes with the direction 
of the motion, we shall have as our expression for the work 
done 

/ F cos Ods. 

Thus if a constant force of 100 lbs. act upon a body in a direc- 
tion making an angle of 30 with the line of motion, then will 
the work done by the force during the time in which it moves 
through a distance of 10 feet be 

(100) (0.86603) (10) = 866 foot-lbs. 



ROTATION OF RIGID BODIES. 



105 



%%6. Rotation of Rigid Bodies. — Suppose a rigid body 
(Fig. 58) to revolve about an axis perpendicular to the plane of 
the paper, and passing through ; 
imagine a particle whose weight is 
w to be situated at a perpendicular 
distance OA = r from the axis of 
rotation, and let the angular velocity 
be a : let it now be required to find 
the moment of the force or forces 
required to impart this motion in a 
unit of time ; for we know, that, if 
the axis of rotation pass through the centre of gravity of the 
body, the motion can be imparted only by a statical couple ; 
whereas if it do not pass through the centre of gravity, the 
motion can be imparted by a single force. 

We shall have, for the particle situated at A, 




Fig. 58. 



Weight = w. 

Angular velocity =: a. 

Linear velocity = a.r. 

Force required to impart this velocity in a unit of time to 

IV 

this particle = — or. 
g 

Moment of this force about the axis — —ar 2 . 

g 

Hence the moment of the force or forces required to impart 
to the entire body in a unit of time a rotation about the axis 
through O, with an angular velocity a, is 



g 



a -%wr 2 = n ^ 



where /is used as a symbol to denote the limit of %wr 2 , and is 
called the Moment of Inertia of the body about the axis through O. 



106 APPLIED MECHANICS. 

§87. Angular Momentum This quantity,—, which ex- 

g 
presses the moment of the force or forces required to impart to 
the body in a unit of time the angular velocity a about the axis 
in question is also called the Angular Momentum of the body 
when rotating with the angular velocity a about the given 
axis. 

§ 88. Actual Energy of a Rotating Body. — If it be required 
to find the actual energy of the body when rotating with the 
angular velocity a, we have, for the actual energy of the particle 
at A, 

W (ar) 2 a 2 



wr z , 



and for that of the entire body 

a 2 ^ a 2 / 

— 2<wr 2 = 

2g 2g 

This is the amount of mechanical work which would have to be 
done to bring the body from a state of rest to the velocity a, or 
the total amount of work which the body could do in virtue 
of its velocity against any resistance tending to stop its 
rotation. 

§ 89. Moment of Inertia. — The term "moment of inertia" 
originated in a wrong conception of the properties of matter. 
The term has, however, been retained as a very convenient one, 
although the conceptions under which it originated have long 
ago vanished. The meaning of the term as' at present used, in 
relation to a solid body, is as follows : — 

The moment of inertia of a body about a given axis is the 
limit of the sum of the products of the weight of each of the ele- 
mentary particles that make up the body, by the squares of their 
distances from the given axis. 

Thus, if Wfy w 2 , w 3 , etc., are the weights of the particles 
which are situated at distances r lt r 2 , r v etc., respectively from 



MOMENT OF INERTIA OF A PLANE SURFACE. lOJ 

the axis, the moment of inertia of the body about the given 

axis is 

/ = limit of Hwr 2 . 

§90. Radius of Gyration. — The radius of gyration of a 
body with respect to an axis is the perpendicular distance from 
the axis to that point at which, if the whole mass of the body 
were concentrated, the angular momentum, and hence the mo- 
ment of inertia, of . the body, would remain the same as they are 
in the body itself. 

If p is the radius of gyration, the moment of inertia would 
be, when the mass is concentrated, 

p z %w ; 
hence we must have 

p 2 %w = %wr 2 = /, 
whence 

Swr 2 I 



2w ~ W 9 



where W = entire weight of the body. 

§91. Moment of Inertia of a Plane Surface.-^- The term 
"moment of inertia," when applied to a plane figure, must, of 
course, be defined a little differently, as a plane surface has no 
weight ; but, inasmuch as the quantity to which that name is 
given is necessary for the solution of a great many questions, 
and also since a knowledge of the manner of determining the 
moments of inertia of plane figures is very useful in simplifying 
the determinations of those of solid bodies, we shall now take 
up those of plane figures. 

The moment of inertia of a plane surface about an axis, either 
in or not in the plane, is the limit of the sum of the products of 
the elementary areas into which the surface may be conceived to 
be divided, by the squares of their distances from the axis in 
question. 



108 APPLIED MECHANICS. 

From this definition it will be evident, that, if the surface be 
referred to a pair of axes in its own plane, the moment of iner- 
tia of the surface about O Y will be 

I=ffx 2 dxdy, (i) 

and the moment of inertia of the surface about OX will be 

J=fffdxdy. (2) 

The moment of inertia of the surface about an axis passing 
through the origin, and perpendicular to the plane XO Y, will be 

ffr 2 dxdy, (3) 

where r= distance from O to the point {x,y)\ hence r 2 = x 2 -{- 
y 2 , and the moment of inertia becomes 

ff(x 2 + y 2 )dxdy = ffx 2 dxdy + ffy 2 dxdy = / + /. (4) 

This is called the "polar moment of inertia." If polar co-ordi- 
nates be used, this last becomes 

ffp 2 (pd P dO) = ffp'dpdO. (5) 

All these quantities are quantities that will arise in the discus- 
sion of stresses, and the letters / and J are very commonly used 
to denote respectively 

ffx 2 dxdy and fffdxdy. 

Another quantity that occurs also, and which will be repre- 
sented by K, is 

ffxydxdy; (6) 

and this is called the moment of deviation. 



EXAMPLES OF MOMENTS OF INERTIA. 



IO9 



EXAMPLES. 



The following examples will illustrate the mode of finding the 
moment of inertia : — x 



1. Find the moment of inertia of the rectangle 



ABCD about OY (Fig. 59). 



< b 



Solution. 



I = C C 2 x 2 dxdy = b f*x 2 dx = ^. 



Fig. 



2. Find the moment of inertia of the entire circle (radius r) about 
the diameter OY (Fig. 60). 



Fig. 60. 



Solution. 




I 



/ x 2 dxdy — 2 I x 2 *Jr> — x 2 dx 

= zj — ± x (r* — x*\ + — C\l^-x 2 dxV 



4 64' 



3. Find the moment of inertia of the circular ring (outside radius r, 
inside radius r x ) about OY (Fig. 61). 



/ = 



Solution. 
7T/-4 irr* ir{r* — rj) ir(d* — dS) 
4 4 4 64 



4. Find the moment of inertia of an ellipse 
(semi-axes a and b) about the minor axis OY. 




Fig. 61. 



10 



APPLIED MECHANICS. 



Solution. 



Equation of ellipse is 



\ia*~. 



f 



l 2+ b 



Iy=\ \ x 2 dxdy 

2b C a /- 
= — J x 2 \a 2 



^-tCt)- 2 ? 



On the other hand, I x = 



■ab* 



§ 92. Moments of Inertia of Plane Figures about Parallel 
Axes. 

Proposition. — The moment of inertia of a plane figure 
about an axis not passing through its centre of gravity is equal 
to its moment of inertia about a parallel axis passing through its 
centre of gravity i7icr eased by the product obtained by multiply- 
ing the area by the square of the distance between the two axes. 

Proof. —Let A BCD 
(Fig. 62) be the surface ; let 
OYbe the axis not passing 
through the centre of grav- 
ity ; let P be an elementary 
area AxAy, whose co-ordi- 
nates are OP = x and RP 
= y; and let 00, = a = a 
constant ' = distance be- 
tween the axes. 
Let V R = x x = abscissa of P with reference to the axis 
passing through the centre of gravity, 

x = a -+- x t 
x 2 — x, 2 -h 2ax t ■+ a 2 
.*. x 2 ^xly = x 2 ±xAy -f- 2axkx\y + tf 2 A#Av. 




POLAR MOMENT OF INERTIA OF PLANE FIGURES. Ill 



Hence, summing, and passing to the limit, we have 

ffx 2 dxdy = ffx 2 dxdy + 2affx 1 dxdy + a 2 ffdxdy; (i) 

but if we were seeking the abscissa of the centre of gravity 
when the surface is referred to Y 1 0Y lf and if this abscissa be 
denoted by x of we should have 

ffx^dxdy < 



X Q — 



ffdxdy 



and, since x Q = o, .*. ffx^dxdy — o; hence, substituting this 
value in (i), we obtain 

ffx 2 dxdy = ffx 2 dxdy -f- a 2 ffdxdy. (2) 

If, now, we call the moment of inertia about OY, I, that 
about X Y x , I„ and let the area = A = ffdxdy, we shall have 

I=I l + a-A. (3) 

Q. E. D. 

§ 93. Polar Moment of Inertia of Plane Figures. — The 

moment of inertia of a plane 
figure about an axis perpen- 
dicular to the plane is equal 
to the stem of its moments 
of inertia about any pair of D J 
rectangular axes in its plane 
passing through the foot of 
the perpendicular. 

Proof. — Let BCD (Fig. 
63) be the surface, and P an 
elementary area, and let 
OA = x, AP = y, OP = r; then the moment of inertia of 
the surface about OZ will be 




Fig. 63. 



ffr 2 dxdy = ff(x 2 + f)dxdy = ffx z dxdy -f ffy 2 dxdy 
Q. E. D. 



/+/■ 



112 APPLIED MECHANICS. 

Hence follows, also, that the sum of the moments of inertia 
of a plane surface relatively to a pair of rectangular axes in its 
own plane is isotropic ; i.e., the same as for any other pair of 
rectangular axes meeting at the same point, and lying in its 
plane. 

EXAMPLES. 

i. To find the moment of inertia of the rectangle (Fig. 59) about 
an axis through its centre perpendicular to the plane of the rectangle. 

Solution. 

Moment of inertia about YY = — , 

12 

Moment of inertia about an axis through its 

centre and perpendicular to YY = — ; 

12 



hence 



Polar moment of inertia = — H = — {h 2 + o 2 ). 

12 12 12 



2. To find the moment of inertia of a circle about an axis through 
its centre and perpendicular to its plane (Fig. 60). 



Solution. 

Moment of inertia about OY = — , 

. 4 

Moment of inertia about OX = — j 

4 



hence 



Polar moment of inertia = (- — = — • 

A A 2 



3. To find the moment of inertia of an ellipse about an axis passin; 
through its centre and perpendicular to its plane. 



MOMENTS OF INERTIA ABOUT DIFFERENT AXES. 113 



Solution. 
From example 4, §91, we have 

Ix — 



/,= 



ra^d 



Polar moment of inertia = - — {a 2 -\- b 2 ). 

4 



§ 94. Moments of Inertia of Plane Figures about Different 
Axes compared. — Given the surface KLM (Fig. 64), suppose 
we have already determined the quantities 

/ = ffx 2 dxdy, J = fffdxdy, K = ffxydxdy, 

it is required to determine, in terms of them, the quantities 

I* = ffx 1 2 dx l dy lJ /, = ffyfdxjy,, K x = ffx I y J dx,dy 1 ; 



the angles XOYznd X x OY, being both 
right angles, and YO Y Y = a. 

We shall have, from the ordinary 
equations for the transformation of co- 
ordinates, to be found in any analytic 
geometry, the equations 



x t = x cos a 4- y sin a, 

y x = y COS a — X sin a, 

j*^ 2 = .# 2 cos 2 a 4- y 2 sin 2 a 4- 2xy cos a sin a, 
j x 2 = .r 2 sin 2 a -f- j 2 cos 2 a — 2xy cos a sin a, 
*i>i = a.j(cos 2 a — sin 2 a) — {x 2 — y 2 ) cos a sin a. 




Fig. 64. 



I 14 APPLIED MECHANICS. 



Hence 



/, = ffx?dxjy l = limit of %x 2 AA 

= cos 2 a limit of 1x 2 AA + sin 2 a limit of tfAA 4- 

2 cos a sin a limit of ~%xyAA 
= (cos 2 a) ffx 2 dxdy 4- (sin 2 a)ffy 2 dxrdy + 

2 (cos a sin a) ffxydxdy. 
J x = ffy x *dx x dy x = limit of 2)' r 2 A,4 

= (sin 2 a) limit of %x 2 AA + (cos 2 a) limit of Sjj^A^ — 

2 (cos a sin a) limit of ^xyAA 
= (sin 2 a)ffx 2 dxdy + ( cos 2 a) ffy 2 dxdy — 

2 (cos a sin a) ffxydxdy. 
K t — ffx 1 y 1 dx 1 dy I = limit of %x x y\AA 

= (cos 2 a — sin 2 a) limit of %xy\A — (cos a sin a) \ limit of 

1x 2 A A — limit of %y 2 A A \ 
= (cos 2 a — sin 2 a) ffxydxdy — (cos a sin a) \ffx 2 dxdy — 

fffdxdy\. 

Or, introducing the letters /,_/", and isf, we have 

I x = /cos 2 a -f- ysin 2 a 4 2J cos a sin a, (1) 

J x = /sin 2 a 4 fcos 2 a — 2^ COS a sin a, (2) 

K, = — (/— /) cos a sin a + A'(cos 2 a — sin 2 a). (3) 

The equations (1), (2), and (3) furnish the solution of the 
problem. 

§95. Principal Moments of Inertia in a Plane. — /;/ every 
plane figure, a given point being as sinned as origin, there is at 
least one pair of rectangular axes, about one of which the moment 
of iiiertia is a maximum, and a minimum about the other ; these 
moments of inertia being called principal moments of inertia, 
and the axes about which they are taken being called principal 
axes of inertia. 



AXES OF SYMMETRY OF PLANE FIGURES. 115 

Proof. — In order that /„ equation (1), §94, may be a maxi- 
mum or a minimum, we must have, as will be seen by differen- 
tiating its value, and putting the first differential co-efficient 
equal to zero, 

— 2/cosasina 4- 2y"cosasina -j- 2jT(cos 2 a — sin 2 a) = o 

.-. K{ cos 2 a — sin 2 a) — (/ — J) cos a sin a = o (1) 

cos a sin a K zK , , 

tan 2a = -. (2) 



cos 2 a — sin 2 a I — J I — J 

Hence, for the value of a given by (2), we have /, a maximum 
or a minimum ; and as there are two values of 2a corresponding 
to the same value of tan 2a, and as these two values differ by 
180 , the values of a will differ by 90 , one corresponding to a 
maximum and the other to a minimum. 

Moreover, when the value of a is so chosen, we have 

JT/= o, 

as is proved by equation (1). Indeed, we might say that the 
condition for determining the principal axes of inertia is 

K x = o. 

§ 96. Axes of Symmetry of Plane Figures. — An axis 
which divides the figure symmetrically is always a principal 
axis. 

Proof. — Let us assume that the y axis divides the surface 
symmetrically ; then we shall have, with reference to this axis, 



K — J I xydydx = j / — 



ydy[ =0. 



And, since K is zero, the axis of y is one principal axis, and of 
course the axis of x is the other. The same method of reason- 
ing would show K = o if the x axis were the axis of symmetry. 



I l6 APPLIED MECHANICS. 



Hence, whenever a plane figure has an axis of symmetry, 
this axis is one of the principal axes, and the other is at 
right angles to it. Thus, for a rectangle, when the axis is to 
pass through its centre of gravity, the principal axes' are par- 
allel to the sides respectively, the moment of inertia being 
greatest about the shortest axis, and least about the longest. 
Thus in an ellipse the minor axis is the axis of maximum, 
and the major that of minimum, moment of inertia, etc. On 
the other hand, in a circle, or in a square, since the maximum 
and minimum are equal, it follows that the moments of inertia 
about all axes passing through the centre are the same. 

§ 97. Conditions for Equal Values of Moment of In- 
ertia. — When the moments of inertia of a plane figure about 
three different axes passing through the same point are the. 
same, the moments of inertia about all axes passing through 
this point are the same. 

Proof. . — Let I be the moment of inertia about O Y, I x 
about O Y tt I 2 about O Y 2 , and let 

YOYt = a, Y0Y 2 = /?, 
and let 

A*= / 2 = /. 

Then, from equation (1), § 94, we have 

1=1 cos 2 a -h J sin 2 a -f- 2 K cos a sin a, 

1=1 cos 2 /3 -f J sin 2 f3 + 2 ^cos]8 sin ft. 



Hence 



Hence 



(/ — y)sin 2 a = 2 A' cos a sin a, (1) 

(/-/)sin*£ = 2 A'cos/3sin/3. (2) 



(7-/)tana = 2AT, (3) 

(I-J)\smP=2K. (4) 

And, since tan a is not equal to tan ft, we must have 
I — J = o and K = o. 

Hence, since K — o and I ' = J, we shall have, from equa- 



MOMENTS OF INERTIA ABOUT PARALLEL AXES. 11/ 

tion (i), § 94, for the moment of inertia / / about an axis, 
making any angle 6 with OY, 

I' = /cos 2 6 + /sin 2 + o = I. (5) 

Hence all the moments of inertia are equal. 

§ 98. Components of Moments of Inertia of Solid 
Bodies. — Refer the body to three rectangular axes, OX, OY, 
and OZ ; and let I M I y , and I z represent its moment of inertia 
about each axis respectively. Then, if r denote the distance of 
any particle from OZ, we shall have 

I z = limit of %wr 2 ; 
but 

r* = x 2 + y 2 

. • . I z = limit of %w{x 2 -f y 2 ) = limit of %wx 2 -f- limit of %wy 2 . (1) 

In the same way we have 

I x = limit of %wy 2 + limit of %wz 2 , (2) 

Iy = limit of Szeu; 2 + limit of ^wz 2 . (3) 

§99. Moments of Inertia of Solids around Parallel 
Axes. — The moment of inertia of a solid body about an axis 
not passing through its centre of gravity is equal to its moment 
of inertia about a parallel axis passing through the centre of 
gravity, increased by the product of the entire weight of the 
body by the square of the distance between the two axes. 

Proof. — Refer the body to a system of three rectangular 
axes, OX, OY, and OZ, of which OZ is the one about which 
the moment of inertia is taken. Let the co-ordinates of the 
centre of gravity of the body with reference to these axes be 
(Xo> y Q y £<>)■ Through the centre of gravity of the body draw a 
system of rectangular axes, parallel respectively to OX, OY, and 
OZ. Then we shall have for the co-ordinates of any point 

X — Xq ~j~ Xi, 

y = jo +J n 

z = Z Q + *,. 



i8 



APPLIED MECHANICS. 



Hence 

I z — limit of ^w{x 2 -f- y 2 ) = limit of 3m* 2 + limit of Hwy 2 
= limit of 2w(x + j^) 2 -h limit of %w(y + y r ) 2 
= x 2 limit of ~%w + j)'o 2 limit of ^w + 2^ limit of %wx t 

+ 2j' limit of %wy l + limit of 2ze/.*i 2 -f limit of 2z£/v t 2 
= (*o 2 + Jo 2 ) ^+ 2^ limit of SzcmCi -f- 2j limit of 2a/y x 

-f- limit of lour? 
= r 2 W + // + 2x limit of 2 t wx I -j- 2y Q limit of S«y x . 

But, since 6^ is the centre of gravity, 

.*. %wx 1 = o and ^wy 1 = o. 
Hence 



/. = h 



Wr n 



which proves the proposition. 

§ ioo. Examples of Moments of Inertia. 

i . To find the moment of inertia of a sphere whose radius is r and 
weight per unit of volume w, about the axis OZ drawn through its centre. 

Solution. 

Divide the sphere into thin slices (Fig. 65) by planes drawn perpen- 
dicular to OZ. Let the distance 
of the slice shown in the figure, 
above O be z, and its thickness dz: 
then will its radius be vV 2 — z 2 ; 
and we can readily see, from ex- 
ample 2, § 
inertia about OZ will be 




§ 93, that its moment of 



Fig. 65. 



wrr{r* — z 2 ) 2 ^ 

2 

Hence the moment of inertia 
of the entire sphere about OZ will 
be 



t/— r 



I z = w- j (r* — 2 2 ) 2 A 



EXAMPLES OF MOMENTS OF INERTIA. 



II 9 



which easily reduces to 



/, 



wn-r 5 . 



15 



2. To find the moment of inertia of an ellipsoid (semi-axes a, b, c) 
about OZ (Fig. 66). 

Solution. — The equa- 
tion of the ellipsoid is 

x 2 , y 2 , z 2 



Divide it into thin slices 
perpendicular to OZ, and s 
let the slice shown in the 
figure be at a distance z 
from O. Then will this 
slice be elliptical, and its 
semi-axes will be 




Fig. 66. 



S/c 



and 



- V^ 2 - z 2 ; 



and from example 3, §93, we readily obtain, for its moment of inertia 
about OZ, 



"&*-*>]&* r *+%*-*)) 



dz 



Hence, for the moment of inertia of the ellipsoid about OZ, we 



» -I- p\ re 

I z s= - ^7- I (c 2 — z 2 ) 2 dz = ^-wirabc{a 2 -f- b 2 ). 

•J— c 



have 

wirab(a 2 4- b 2 ) 1 

— L — / (^ 2 — s 2 )Vz = -L 

3. Find the moment of inertia of a right circular cylinder, length a, 
radius r, about its axis. 

WTT7' A a 
2 



Ans. 



120 APPLIED MECHANICS. 

4. Find the moment of inertia of the same about an axis perpen- 
dicular to, and bisecting its axis. 

. wrrar 2 [ , , a 2 \ 
Ans. [ r 2 H ). 

4 V 3/ 

5. Find the moment of inertia of an elliptic right cylinder, length 
2<r, transverse semi-axes a and b } about its longitudinal axis. 

Ans. (a 2 -f b 2 ). 

2 

6. Find the moment of inertia of the same about its transverse 
axis 2b. 

Ans. 2Wirabcl 1 ]. 

1 4 3/ 

7. Find the moment of inertia of a rectangular prism, sides 2a, 2b, 
2C, about central axis 2c. 

o 

Ans. -wabc{a 2 -f- b 2 ). 
3 

§101. Centre of Percussion. — Suppose we have a body 
revolving about an axis perpendicular 
to the plane of the paper, and passing 
through O, with an angular velocity a. 
If, with O as a centre and a radius 
OA = r, we describe an arc CB (Fig. 
67), all particles situated in this arc 
have a linear velocity a.r. The mea- 
FlG<67 " sure of the force which would impart 

this velocity to the particle in a unit of time, is 

w 
- or, 

g 

and the moment of this force about the axis is 

g 
hence the total angular momentum, or the total moment of the 




CENTRE OF PERCUSSION. 121 



forces which would impart to the body in a unit of time the an- 
gular velocity a, is, as has been shown already, 

g S 
The sum of the forces acting on the body is, on the other hand, 

a v 

g 

Hence the perpendicular distance from O to the line of direc- 
tion of the resultant force is 

, = _il=JL. (I) 

— 'Xwr 
g 

and if a line be drawn from O in the plane of the paper, perpen- 
dicular to the direction of the resultant, and the length /, as 
deduced above, laid off, the resultant force may be conceived to 
have its point of application at this point, and this point of 
application of the resultant of the forces which produce the 
rotation is called the Centre of Percussion. 

If p = radius of gyration about the axis through O, and if 
r = distance from O to the centre of gravity, we have 

roXw = lour. 
Hence 

%wr 2 I i / I\ p 2 

2,wr r 2,w r \lV/ r Q 

or, in words, — 

The radius of gyration is a mean proportional between the 
distance\,and the distance r Q between' the axis of oscillation and 
the centre of gravity. 

The centre of percussion with respect to a given centre of 
oscillation O has been defined as the point of applicatiojt of the 



122 APPLIED MECHANICS. 

resultant of the forces which cause the body to rotate around the 
point O. 

Another definition often given is, that it is the point at which, 
if a force be applied, there will be no shock on the axis of oscilla- 
tion ; and these two definitions are equivalent to each other. 

Let the particles of the body under consideration be con- 
ceived, for the sake of simplicity, to be distributed along a single 
line AB, and suppose a force F applied at D 
(Fig. 68). Conceive two equal and opposite 
forces, each equal to F, applied at C, the cen- 
tre of gravity of the body. 

I Then these three forces are equivalent to 

a single force ^applied at the centre of grav- 
ity C, which produces translation of the whole 
body ; and, secondly, a couple whose moment 
is F(CD), whose effect is to produce rotation 
Fig. 68. around an axis passing through the centre of 

gravity C. Under this condition of things it is evident that the 

centre of gravity C will have imparted to it in a unit of time a 

p 
forward velocity equal to — , where M is the entire mass of the 

body ; the point D will have imparted to it a greater forward 
velocity ; while those points on the upper side of C will have 
imparted to them a less and less velocity as they recede from 
C, until, if the rod is sufficiently long, the particle at A will 
acquire a backward velocity. 

Hence there must be some point which for the instant in 
question is at rest; i.e., where the velocity due to rotation is just 
equal and opposite to that due to the translation, or about 
which, for the instant, the body is rotating: and if this point 
were fixed by a pivot, there would be no stress on the pivot 
caused by the force applied at D. 

An axis through this point is called the Instantaneous 
Axis. 



impact or collision: 123 

§ 102. Interchangeability of the Centre of Percussion 
and Centre of Oscillation. — If we take our centre of percus- 
sion D as axis of oscillation, then will O be the new centre of 
percussion. 

Proof. — We have seen (§ 101) that 

>-% 

where / = OD, r Q = OC, and p = radius of gyration about an 
axis through O perpendicular to the plane of the paper. 

Moreover, if p Q represent the radius of gyration about an 
axis through C perpendicular to the plane of the paper, we shall 
have 

P 2 = Po 2 + ^o 2 

r Q 
■\ / - r = ^ = CD. 

Now if D is taken as axis of oscillation, we shall have for the 
distance / x to the corresponding centre of percussion, 

h = 



CD I - r Q 7 

where Pl = radius of gyration about the axis of oscillation 
through D. 

Hence the new centre of percussion is at O. Q. E. D. 

§ 103. Impact or Collision. — Impact or collision is a 
pressure of inappreciably short duration between two bodies. 

The direction of the force of impact is along the straight line 
drawn normal to the surfaces of the colliding bodies at their 
point of contact, and we may call this line the line of impact. 



124 APPLIED MECHANICS. 

The action that occurs in the case of collision may be de- 
scribed as follows : at first the bodies undergo compression ; 
the mutual pressure between them constantly increasing, until, 
when it has reached its maximum, the elasticity of the mate- 
rials begins to overpower the compressive force, and restore 
the bodies wholly or partially to their original shape and dimen- 
sions. 

Central impact occurs when the line joining the centres of 
gravity of the bodies coincides with the line of impact. 

Eccentric impact occurs when these lines do not coincide. 

Direct inrpact occurs when the line along which the relative 
motion of the bodies takes place, coincides with the line of 
impact. 

Oblique impact occurs when these lines do not coincide. 

CENTRAL IMPACT. 

§ 104. Equality of Action and Re-action. — One funda- 
mental principle that holds in all cases of central impact is the 
equality of action and re-action ; in other words, we must have, 
that, at every instant of the time during which the impact is 
taking place, the pressure that one body exerts upon the other 
is equal and opposite to that exerted by the second upon the 
first. 

The direct consequence of this principle is, that the algebraic 
sum of the momenta of the two bodies before impact remains 
unaltered by the impact, and hence that this sum is just the 
same at every instant of, and after, the impact. 

If we let 

m„ m 2 , be the respective masses, 
c n c 2 , their respective velocities before impact, 

v 1} v 2 , their respective velocities after impact, 

v\ v" } their respective velocities at any given instant during 
the time while impact is taking place, 



CO-EFFICIENT OF RESTITUTION. 12$ 

then we must have the following two equations true ; viz., — 
m^ + m 2 v 2 = m 1 c i -f- w 2 ^) (i) 

z/^z/ + m*v" = zw^, 4- m 2 c 2 . (2) 

§ 105. Velocity at Time of Greatest Compression. — At 

the instant when the compression is greatest — i.e., at the 
instant when the elasticity of the bodies begins to overcome 
the deformation due to the impact, and to tend to restore them 
to their original forms — the values of v' and v" must be equal 
to each other; in other words, the colliding bodies must be 
moving with a common velocity 

v = ff = v ". (1) 

To determine this velocity, we have, from equation (2), § 104, 
combined with (1), 

v _ m t c s + m 2 c^ ^ 

m Y + m 2 

§ 106. Co-efficient of Restitution. — In order to determine 
the values v iy v 2J of the velocities after impact, we need two 
equations, and hence two conditions. One of them is fur- 
nished by equation (1), § 104. The second depends upon the 
nature of the material of the colliding bodies, and we may dis- 
tinguish three cases : — 

i°. Inelastic Impact. — In this case the velocity lost up to 
the time of greatest compression is not regained at all, and 
the velocity after impact is the common velocity v at the instant 
of greatest compression. In this case the whole of the work 
used up in compressing the bodies is lost, as none of it is 
restored by the elasticity of the material. 

2°. Elastic Impact. — In this case the velocity regained 
after the greatest compression, is equal and opposite to that 
lost up to the time of greatest compression ; therefore 

V — V t = € x — V. (i) V 2 — V = V — C 2 . (2) 



126 APPLIED MECHANICS. 

We may also define this case as that in which the work lost 
in compressing the bodies is entirely restored by the elasticity 
of the material, so that 

1 — 1 • (3) 

2 2 2 2 

Either condition will lead to the same result. 

3°. Imperfectly Elastic Impact. — In this case a part only 
of the velocity lost up to the time of greatest compression is 
regained after that time. 

If, when the two bodies are of the same material, we call e 
the co-efficient of restitution, then we shall so define it that 

v — v 1 v 2 — V 



C 1 — V V — c 2 

or, in words, the co-efficient of restitution is the ratio of the 
velocity regained after compression to that lost previous to 
that time. 

In this case only a part of the work done in producing the 
compression is regained, hence there is loss of energy. Its 
amount will be determined later. 

Strictly speaking, all bodies belong to the third class ; the 
value of e being always a proper fraction, and never reaching 
unity, the value corresponding to perfect elasticity ; nor zero, 
the value corresponding to entire lack of elasticity. 

§ 107. Inelastic Impact. — In this case the velocity after 
impact is the common velocity at the time of greatest com- 
pression ; hence 

V — v v — v 2 (1) 

... v = m ^ + m *\ ( 2 ) 

m l -j- m 2 

And for the loss of energy due to impact we have 

_J_L_ _| LiL_ _ (; ;/j _}_ Ph ) _ 

2 2 2 



ELASTIC IMPACT. 12? 



which, on substituting the value of v, reduces to 

m{m 2 / \ 2 / \ 

— L - J r (*, - c 2 )\ (3) 

2{m l H- m 2 ) 

§ 108. Elastic Impact. — In this case we have, of course, 
the condition, equation (1), § 104, 

ntjVj -+- m 2 v 2 = wi I c l -f- m 2 c 2 , 

and, for second equation, we may use equation (3), § 106; viz., 

m z v? m 2 v 2 _ m x c^ m 2 c 2 * 



Combining these two equations, we shall obtain 

2m 2 (c l — c z ) , N 

v x = c x ^ 2J -, (1) 

m 1 + m 2 

. 2tnAc x — c 2 ) , x 

v 2 = c 2 -\ LV-S- 2 -l. (2) 

m x 4- tn 2 

We can obtain the same result without having to solve an 
equation of the second degree, by using instead the equations 
(1) and (2) of § 106, together with (1) of § 104; i.e., — 

m l v l 4- w 2 ^2 = m^i -f- m z c z '} 

v — v x = c x — v, 
or 

V 2 — V = V — c 2i 
and (§ 105) 

m l c I 4- m 2 c 2 



v 



m 1 4- ^ 2 



As the result of combining these equations, and eliminating 
v, we should obtain equations (1) and (2), as above, for the values 
of v l and v 2 . In this case the energy lost by the collision is 
zero. 



128 APPLIED MECHANICS. 

§ 109. Special Cases of Inelastic Impact. — (a) Let the 

mass m 2 be at rest. Then c 2 = o, 

(*) 



m l -f- 7n 2 







VI 


1 + m z 










of 


energy 


= 


m x 


2 


= 


m x c^ 


2 


2i + 1 


2 





T r 771,7712 Cr / v 

.*. Loss of energy = - — -. (2) 

m l -\- in 2 2 

(3) Let m 2 be at rest, and let m 2 = 00 ; i.e., let the mass m x 
strike against another which is at rest, and whose mass is in- 
finite. We have 

?n 2 = 00 , c 2 — o, 

* = 7— rr = °> (3) 

(4) 

+ 1 2 

or the moving body is reduced to rest by the collision, and all 
its energy is expended in compression. 

(c) Let m l c 1 = —m 2 c 2 ; i.e., let the two bodies move towards 
each other with equal momenta : 

/. y = »* + m * c * = o, (s) 

^1 + ^2 

and the loss of energy = ' * -\ ^- J -, (6) 

2 2 

the entire energy being lost. 

§110. Special Cases of Elastic Impact. — (#) Let the 
mass m 2 be at rest. Then c 2 = o, 

27?l 2 C, / \ 

v x = c x ~— (1) 

»/, -f w 2 

.-. * 3 = -^-. (a) 



EXAMPLES OF ELASTIC AND INELASTIC IMPACT. 120, 

(b) Let m 2 be at rest, and let m 2 == 00 . Then we have 
c , = o, 

.-. Vl = C x = C x — 2C X - —C l9 (3) 

m 2 

v 2 = o. (4) 

Hence the moving body retraces its path in the opposite direc- 
tion with the same velocity. 

(c) Let m 1 c 1 = — m 2 c 2 . Then our equations of condition 

become 

m 1 v 1 + m 2 v 2 = o, 

2222 
and from these we readily obtain 

v t = — c v 

v 2 = —c 2 \ 

i.e., both bodies return on their path with the same velocity 
with which they approached each other. 

§111. Examples of Elastic and of Inelastic Impact. 

1. With what velocity must a body weighing 8 pounds strike one 
weighing 25 pounds in order to communicate to it a velocity of 2 feet 
per second, {a) when the bodies are perfectly elastic, (b) when wholly 
inelastic. 

2. Suppose sixteen impacts per minute take place between two bodies 
whose weights are respectively 1000 and 1200 pounds, their initial velo- 
cities being 5 and 2 feet per second respectively : find the loss of energy, 
the bodies being inelastic. 

§ 112. Imperfect Elasticity. — In this case we have the 
relations (see § 106) 

V — V Y _ v 2 — V __ 
C l — V V — c 2 



I30 APPLIED MECHANICS. 



where 

n 
v — - 

m t -h tn-. 



m,c, -f- m 2 c 2 
v = - 



and we have also 

771 J! x 4" ^2^2 = ^I^I + ^2^2' 

Determining from them the values of v x and v 2 , we obtain 
v s = z/(i + e) — ec Zi (1) 

v 2 = v(i + e) - ec z ; (2) 

or, by substituting for v its value, 

Vi = ; — — (1 + - "« (3) 

w x 4- w 2 

^2 = ; — — (1 4- *) — ^ 2 . (4) 

These may otherwise be put in the form 

Vl = ,, _ (1 + ,) _^_ (,, _ , 2 ), (5) 
«, 4- w 2 

v 2 = c 2 4- (1 4- <?) — v (c t - c 2 ). (6) 

;«! 4- 77i 2 

Moreover, we have for the loss of energy due to impact 
or 



E = 7 -^{c 2 - v 2 ) + ^{c 2 2 - v 2 ) 
2 2 



E = Wm^ — »,)(<:, 4- v,) + 0* 2 (r a — v 2 ){c 2 4- * a )j ; 
but, from (5) and (6) respectively, 

« -- (1 + g)w 2 (f, - c 2 ) 

W, + 0* a 

. ._ (1 + e)M t (c t - c 2 ) 

C 2 c/ 2 — 

;//, 4- Mi 



IMPERFECT ELASTICITY. 131 

^ E = (1 + *)fo ~ O J;;,^^ + ^) _ ;; ?iW2 (r 2 + Vz )\ 
2{7?i l -f- m 2 ) 

... E = t m * m * (c x - c 2 )(i + *)(<:, _,*, + *,- z> 2 ). 

2(w, + ;;/ 2 ) 

But, from (1) and (2), 

#1 — #2 = — *(*"i ~~ r a) 

••• ^ = . g ''"' 2 . (*, - *)0 + *)fo - 0(i - 

2(;;/ x + m 2 ) 



or 



^ = (1 - ^) "■"» < fl - ^.. (7) 

2(#z I + m 2 ) 



When e = 1, or the elasticity is perfect, this loss of .energy 
becomes zero. 

When e = o, or the bodies are totally inelastic, then the loss 
of energy becomes 



2(?/7 r + m 2 ) 



('. - t,Y. (8) 



as has been already shown in § 107. 

An interesting fact in this connection is, that since (8) is 
the work expended in producing compression, and (7) is the 
work lost in all, therefore the work restored by the elasticity of 
the body is 



A^^ic-M; (9) 



so that e 2 , or the square of the co-efficient of restitution, is the 
ratio of the work restored by the elasticity of the bodies, to 
the work expended in compressing the bodies up to the time 
of greatest compression. 



132 APPLIED MECHANICS. 

§113. Special Cases. — (a) Let m 2 be at rest, therefore 
c 2 = o. Then we shall have 

v^cAi- »s£l±A\= Cl >i^=j^ f (I) 

\ m 1 ^- m 2 ) m l + m 2 

v z = (1 + e)c t ^ , (2) 

and for loss of energy 

E= (1 __,*) *■"»» ,,.. (3) 

2(/W, -f- /0 2 ) 

(£) When m 2 = 00 , and r 2 = o, we have 

Pi = — '*„ (4) 

V 2 = O, 

i -'('1 - <*) ^. (s) 

2 

(<:) When *«,£, = —m 2 c 2 , then 

^i = -^1, 
z> 2 = — ec 2 , 

£ — (1 — e^m.c^c, — c 2 ) = (1 — e 2 )m 2 c 2 (c 2 — c t ) 
2 2 

= (I _.«,)*■("'« + "0,,,. (6) 

2Vl 2 

§114. Values of e as Determined by Experiment.— 
Since we have 

e, — v 






IMPERFECT ELASTICITY. 1 33 

we shall have, when 

;;/ 2 = oo and c 2 — o, 

v = — — — ! — = o. 

Hence 

- ■ 2i 

Now, if we let a round ball fall vertically upon a horizontal 
slab from the height H, we shall have for the velocity of ap- 
proach 

and if we measure the height h to which it rises on its rebound, 
we shall have 

— Z>! = \2gk. 

Hence 



--*-vi 



In this way the value of e can be determined experimentally 
for different substances. 

Newton found for values of e: for glass, \\ ; for steel, |; 
and Coriolis gives for ivory from 0.5 to 0.6. 

On the other hand, if we desired to adopt as our constant 
the ratio of the work restored, to the work spent in compres- 
sion, we should have for our constant e 1 i and hence the squares 
of the preceding numbers. 



EXAMPLES. 

1. If two trains of cars, weighing 120000 and 160000 lbs., come 
into collision when they are moving in opposite directions with veloci- 
ties 20 and 15 feet per second respectively, what is the loss of mechan- 
ical effect expended in destroying the locomotives and cars ? 



134 APPLIED MECHANICS. 

2. Two perfectly inelastic balls approach each other with equal 
velocities, and are reduced to rest by the collision ; what must be the 
ratio of their weights? 

3. Two steel balls, weighing 10 lbs. each, are moving with velocities 
5 and 10 feet per second respectively, and in the same direction : find 
their velocities after impact, the fastest ball being in the rear, and over- 
taking the other ; also the loss of mechanical effect due to the impact, 
assuming e = 0.5,5. 

§115. Oblique Impact. 

Let m l} m 2 , be the masses of the colliding bodies ; 
c„ c 2} their respective velocities before impact ; 
a ir a 2 , the angles made by r x , c 2 , with the line of centres ; 
v s , v 2y the components- of the velocities after impact ; 
c l cosa iy c 2 co$a 2 , the components of c„ c 2> along the line of 

centres ; 
c x sin a lt c 2 sin a 2 , the components of c u c 2 , at right angles to 

the line of centres ; 
v the common component of the velocity at the instant of 

greatest compression along line of centres ; 
v\ v'\ actual velocities after impact ; 
a, a", angles they make with line of centres ; 
v/j v" f actual velocities when compression is greatest ; 
a/, a/', angles they make with line of centres. 

Then we shall have, by proceeding in the same way as was done 

in §112, 

Vt = ^cos^ — (1 + e) — (^cosa! — <r 2 cosa 2 ), (1) 

m l -\- m 2 



v 2 — C 2 COS a 2 + (1 + e) ■ (^COSd! — r 2 cosa 2 ), (2) 

m l -f- m 2 



OBLIQUE IMPACT. 1 35 



v' = 


V^i 2 + C1 2 sin 2 a„ 


(3) 


z/' = 


V^ 2 2 + ^ 2 2 sin 2 a 2 , 


(4) 


COS a' = 


z/' 


(5) 


COS a" = 


^2 

z/" * • 


(6) 


V = 


tn t e t cos 0, + m 2 ? a cos a. 


(7) 


w, + m 2 


»/.;- 


\/v* + c, 2 sili 2 a„ 


(8) 



')" = vV + c* sin 2 a 2 , (9) 

COSa c = — -, (IO) 

C0Sa ^ = 3' ( IX ) 

And for the energy lost in impact, we have 

E = (1 — e 2 )— ^ — --(^cosa! — <r 2 cosa 2 ) 2 . (12) 

When the bodies are perfectly elastic, 

e = 1, 
and equations (1), (2), and (12) become respectively 

1711 
V x = C x COS a x - (c x COS a x — <T 2 COS a 2 ), 

Off) 

V 2 — C 2 COS a 2 H — Uj, COS a x — <r 2 COS a 2 ), 

»?! -f- #2 2 

E = 0. 

The rest remain the same in form. 

When the bodies are totally inelastic, 

e = o, 



136 APPLIED MECHANICS. 

and equations (1), (2), and (12) become respectively 

V x — f , COS a, — (^, COS a! — <: 2 C0Sa 2 ), 

m L + w 2 

V 2 = C 2 COS a 2 -\ U — (<r x COS a x — C 2 COS a 2 ), 

;;/ x + m 2 

i? = *—2 (c t COS a x — C 2 COS a 2 ) 2 . 

2 (;«i + m 2 ) 
The rest remain the same in form. 

§116. Impact of Revolving Bodies. — Let the bodies A 
and B revolve about parallel axes, and impinge upon each other. 
Draw a common normal at the point of contact. This 
common normal will be the line of impact. 
Let e x — angular velocity of A before impact, 
€ 2 = angular velocity of B before impact, 
(o r = angular velocity of A after impact, 
w 2 = angular velocity of B after impact, 
a z = perpendicular from axis of A on line of impact, 
a 2 = perpendicular from axis of B on line of impact, 
/, = moment of inertia of A about its axis, 
I 2 = moment of inertia of B about its axis. 

Then we shall have 

a^ = Ci = linear velocity of A at point of contact before impact ; 

a 2 € 2 = c 2 = linear velocity of B at point of contact before impact ; 

a 1 o) 1 = v s = linear velocity of A at point of contact after impact ; 

a 2 u> 2 = v 2 = linear velocity of B at point of contact after impact ; 

/ e 2 / 1 \c 2 

— — = f — - )— = actual energy of A before impact ; 

-*-*- = ( —2- )— = actual energy of B before impact ; 
2g W/ 2 A r 

-^- = (— L )— = actual energy of A after impact; 
2g W/ 2 A r 

- 1 ^ ? - = ( — — )— = actual energy of B after impact ; 
ig W/ 2£ 



IMPACT OF REVOLVING BODIES. 1 37 

Hence it follows that we have the case explained in § 112 for 
imperfectly elastic impact, provided only we write 

/ j 

— - instead of m^g and — instead of m 2 g. 
a* a 2 2 

Hence we shall have 

<«>i = €i ~ a x {a z € x — a 2 € 2 )- f— — (1 + e), (1) 

<o 2 = c 2 + tf 2 0i€i — a 2 € 2 )- -±— — (1 -f- e), (2) 

l z a 2 + I 2 a 2 2 

The case of perfect elasticity is obtained by making e == I. 
The case of total lack of elasticity is obtained by making 
e — o. 

In the latter case the loss of energy is 



(a^ — a 2 <: 2 y IJ 2 

2 A#i 2 + / 2 ^ 2 2 ' 



(3) 



as can be seen by substituting the proper values in equation (8), 
§ 112. 



I38 APPLIED MECHANICS. 



CHAPTER III. 
ROOF-TRUSSES. 

§ 117. Definitions and Remarks. — The term "truss" may 
be applied to any framed structure intended to support a load. 

In the case of any truss, the external loads may be applied 
only at the joints, or some of the truss members may support 
loads at points other than the joints. 

In the latter case those members are subjected, not merely 
to direct tension or compression, but also to a bending-action, 
the determination of which we shall defer until we have studied 
the mode of ascertaining the stresses in a loaded beam ; and 
we shall at present confine ourselves to the consideration of 
the direct stresses of tension and compression. 

For this purpose any loads applied between two adjacent 
joints must be resolved into two parallel components acting at 
those joints, and the truss is then to be considered as loaded at 
the joints. By this means we shall obtain the entire stresses in 
the members whenever the loads are concentrated at the joints ; 
and, when certain members are loaded at -other points, our re- 
sults will be the direct tensions and compressions of these mem- 
bers, leaving the stresses due to bending yet to be determined. 

A tie is a member suited to bear only tension. 

A strict is a member suited to bear compression, 

§ 118. Frames of Two Bars. — Frames of two bars may 
consist, (1) of two ties (Fig. 69), (2) of two struts (Fig. 70), 
(3) of a strut and a tie (Fig. 71). 



FRAMES OF TWO BARS. 



139 



Case I. Two Ties (Fig. 69). — Let the load be repre- 
sented graphically by CF = W. 
Then if we resolve it into 

two components, CD and CE, \ -<*?> 

acting: alon°- CB and CA re- 

o o 

spectively, CD will represent 
graphically the pull or tension 
in the tie CB, and CE that in 
the tie CA. 

The force acting on CB at 
B is equal and opposite to 
CD, while that acting on CA at A is equal and opposite to CE. 

To compute these stresses analytically, we have 




Fig. 69. 



CE = CF 



CD = CF 



sin C#g 
sin CEF 

sin 67^£> 
sin CDF 



= JF 



sin z 



sin(* + /,)' 



W 



sm*! 



Case II. 



sin(Y -f- z\) 

Two Struts (Fig. 70). — Let the load be repre- 
sented graphically by CF= W. 
Then will the components CD 
and CE represent the thrusts 
in the struts CB and CA re- 
spectively, and the re-actions 
of the supports at B and A 
will be equal and opposite to 
them. For analytical solution, 
we derive from the figure 




CE 



W 



sin z, 



sin(/ -f t L ) 



CD = W 



sm 1 



sin(/ -h i\) 



Case III. A Strut and a Tie (Fig. 71). — Let the load be 
represented graphically by CF = W. Resolve it, as before, 
into components along the members of the truss. Then will 




140 APPLIED MECHANICS. 

CE represent the tension in the tie AC, and CD will represent 
the thrust in the strut BC ; and we may 
deduce the analytical formulae as before. 

§ 119. Stability for Lateral Deviations.. 
. — In Case I, if the joint C be moved a little 
out of the plane of the paper, the load at 
C has such a direction that it will cause the 
truss to rotate around AB so as to return to 
its former position ; hence such a frame is 
stable as regards lateral deviations. 

In Case II the effect of the load, if C 
were moved a little out of the plane of the 
paper, would be to cause rotation in such a way as to overturn 
the truss ; hence such a frame is unstable as regards lateral 
deviations. 

In Case III the stability for lateral deviations will depend 
upon whether the load CF = W is parallel to AB, is directed 
away from it or towards it. If the first is the case (i.e., if A is 
the point of suspension of the tie), the frame is neutral, as the 
load has no effect, either to restore the truss to its former posi- 
tion, or to overturn it ; if the second is the case (i.e., if A x is 
the point of suspension of the tie), the truss is stable ; and, if 
the third is the case (i.e., if A 2 is the point of suspension of the 
tie), it is unstable as regards lateral deviations. 

§ 120. General Methods for Determining the Stresses in 
Trusses. — In the determination of the stresses as above, it 
would have been sufficient to construct only the triangle CFD 
by laying off CF = IV to scale, and then drawing CD parallel 
to CB, and FD parallel to CA, and the triangle CFD would have 
given us the complete solution of the problem. Moreover, the 
determination of the supporting forces of any truss, and of the 
stresses in the several members, is a question of equilibrium. 
Adopting the following as definitions, viz., — 
External forces are the loads and supporting forces, 



TRIANGULAR FRAME. I4I 

Internal forces are the stresses in the members : 
we must have 

i°. The external forces must form a balanced system; i.e., 
the supporting forces must balance the loads. 

2°. The forces (external and internal) acting at each joint 
of the truss must form a balanced system ; i.e., the external 
forces (if any) at the joint must be balanced by the stresses in 
the members which meet at that joint. 

3 . If any section be made, dividing the truss into two parts, 
the external forces which act upon that part which lies on one 
side of the section, must be balanced by the forces (internal) 
exerted by that part of the truss which lies on the other side 
of the section, upon the first part. 

The above three .principles, the triangle, and polygon of 
forces, and the conditions of equilibrium for forces in a plane, 
enable us to determine the stresses in the different members 
of roof and bridge trusses. 

§121. Triangular Frame. — Given the triangular frame 
ABC (Fig. 72), and given the load W at C in magnitude and 
direction, given also the 
direction of the support- 
ing force at B, to find the 
magnitude of this support- 
ing force, the magnitude 
and direction of the other 
supporting force, and the 
stresses in the members. 

Solution. — Join A 
with D, the point of inter- Fig. 72. 

section of the line of direction of the load and the line BE. 
Then will DA be the direction of the other supporting force ; 
for the three external forces, in order to form a balanced sys- 
tem, must meet in a point, except when they are parallel. 
Then draw ab to scale, parallel to CD and equal to W. From 




142 



APPLIED MECHANICS. 



a draw ac parallel to BD, and from b draw be parallel to AD ; 
then will the triangle abca be the triangle of external forces, 
the sides ab, be, and retaken in order, representing respectively 
the load W, the supporting force at A, and the supporting force 
at B. 

Then from a draw ad parallel to BC, and from c draw cd 
parallel to AB ; then will the triangle acd be the triangle of 
forces for the joint B, and the sides ca, ad, and dc, taken in 
order, will represent respectively the supporting force at B, the 
force exerted by the bar BC at the point B, and the force 
exerted by the bar AB at the point B. 

Since, therefore, the force ad exerted by the bar CB at B 
is directed away from the bar, it follows that CB is in compres- 
sion ; and, since the force dc exerted by the bar AB at B is 
directed towards the bar, it follows that AB is in tension. 

In the same way bde is the triangle of forces for the point 
A ; the sides be, cd, and db representing respectively the sup- 
porting force at A, the force 
exerted by the bar AB at A, 
and the force exerted by the 
bar AC at A. 

The bar AB is again seen to 
be in tension, as the force cd 
exerted by the bar AB at A is 
directed towards the bar. 

So likewise the triangle abd 
is the triangle of forces for the 
point C. 

Fig. 73 shows the case when 
the supporting forces meet the load-line above, instead of 




below, the truss. 

§ 122. Triangular Frame with Load and Supporting 
Forces Vertical. — Fig. 74 shows the construction when the 
load and also the supporting forces are vertical. In this case 



BOW'S NOTATION. 



143 



the diagram becomes very much simplified, the triangle of 
external forces abd becom- 



ing a 



straight line. 




of 



Fig. 74. 

the stress diagrams of roof- 



The 

diagram is otherwise con- 
structed just like the last 
one. 

§ 123. Bow's Notation. 
— The notation devised by 
Robert H. Bow very much 
simplifies the construction 
trusses. 

This notation is as follows : Let the radiating lines (Fig. 75) 
represent the lines of action of a system of forces in equilib- 
rium, and let the polygon abcdefa be the polygon representing 

these forces in magnitude 
and direction ; then denote 
the sides of the polygon 
in the ordinary way, by 
placing small letters at the 
vertices, but denote the 
radiating lines by capital 
letters placed in the angles. 
Thus the line AB is the 
line of direction of the 
force a&, etc. In applying the notation to roof-trusses, we letter 
the truss with capital letters in the spaces, and the stress dia- 
gram with small letters at the vertices. If, then, in drawing 
the polygon of equilibrium for any one joint of the truss, we 
take the forces always in the same order, proceeding always 
in right-handed or always in left-handed rotation, we shall be 
led to the simplest diagrams. Hereafter this notation will be 
used exclusively in determining the stresses in roof-trusses. 

§124. Isosceles Triangular Frame: Concentrated Load 
(Fig. 76.) — Let the load W act at the apex, the supporting 




Fig. 75. 



144 



APPLIED MECHANICS. 




Fig. 76. 



forces being vertical ; each will be equal to % W : hence the 
polygon of external forces will be the triangle abc, the sides of 

which, ab, be, and ca, all lie in 
one straight line. Then begin 
at the left-hand support, and 
proceed again in right-handed 
rotation, and we have as the tri- 
angle of forces at this joint cad, 
the forces ca, ad, and dc, these 
being respectively the support- 
ing force, the stress in AD, and 
that in DC; the directions of 
these forces being indicated by 
the order in which the letters follow each other : thus, ca is an 
upward force, ad is a downward force ; and, this being the 
force exerted by the bar AD at the left-hand support, we con- 
clude that the bar AD is in compression. Again : dc is 
directed towards the right, or towards the bar itself, and hence 
the bar DC is in tension. The triangle of forces for the other 
support is bed, and that for the apex abd. 

§125. Isosceles Triangular Frame : Distributed Load. — 
Let the load W be uniformly distributed over the two rafters 
AF and FB (Fig. 77) ; then will- 
these two rafters be subjected to 
a direct stress, and also to a bend- 
ing action : and if we resolve the 
load on each rafter into two com- 
ponents at the ends of the rafter, 
then, considering these components 
as the loads at the joints, we shall 
determine correctly by our diagram the direct stresses in all 
the bars of the truss. 

The load distributed over AF is — ; and of this, one-half is 




POLYGONAL FRAME. 



145 



the component at the support, and one-half at the apex, and 

similarly for the other rafter. This gives as our loads, — at 

4 

W 
each support, and — at the apex. The polygon of external 

forces is eabcde, where the sides are as follows : — 



W 



ab = 



W 



be = 



W 



cd = 



W 



de = 



W 



Then, beginning at the left-hand support, we shall have for the 

polygon of forces the quadrilateral deafd, where de — — = sup- 

2 

W 
porting force, ea — — = downward load at support, af = 

4 
stress in AF (compression), fd == stress in FD (tension). The 
polygon for the apex is abf, and that for the right-hand support 
cdfbc. 

§ 126. Polygonal Frame Given a polygonal frame (Fig. 

78) formed of bars jointed together at the vertices of the angles, 
and free to turn on these joints, 
it is evident, that, in order that 
the frame may retain its form, 
it is necessary that the direc- 
tions of, and the proportions 
between, the loads at the dif- 
ferent joints, should bt speci- 
ally adapted to the given form : 
otherwise the frame will change 
its form. We will proceed to 
solve the following problem : 
Given the form of the frame, 
the magnitude of one load as AB, and the direction of all the 
external forces (loads and supporting forces) except one, we 
shall have sufficient data to determine the magnitudes of all, 




146 



APPLIED MECHANICS. 



and the direction of the remaining external forces, and also the 
stresses in the bars 

Let the direction of all the loads be given, and also that of 
the supporting force EF, that of the supporting force AF being 
thus far unknown; and let the magnitude of AB be given. 
Then, beginning at the joint ABG, we have for triangle of 
forces abg formed by drawing ab || and = AB, then drawing 
ga || AG, and bg || BG ; ga and bg both being thrusts. Then, 
passing to the joint BCG, we have the thrust in BG already- 
determined, and it will in this case be represented by gb. If, 
now, we draw be || BC, and gc || GC, we shall have determined 
the load BC as be, and we shall have eg and gb as the thrusts 
in CG and GB respectively. Continuing in the same way, we 
obtain the triangles gcd, gde, and gfe, thus determining the 
magnitudes of the loads cd, de, and of the supporting force ef; 
and then the triangle gaf, formed by joining a and/, gives us af 
for the magnitude and direction of the left-hand support. The 
polygon abedefa of external forces is called the Force Polygon, 
while the frame itself is called the Equilibrium Polygon. 

§ 127. Polygonal Frame with Loads and Supporting 
Forces Vertical. — In this case (Fig. 79) we may give the 

form of the frame and the mag- 
nitude of one of the loads, to 
determine the other loads and 
the supporting forces, and also 
the stresses in the bars ; or we 
may give -the form of the frame 
and the magnitude of the re- 
sultant of the loads, to find the 
loads and supporting forces. In 
the former case let the load AB 
be given. Then, proceeding in 
the same way as before, we find the diagram of Fig. 79 ; the 
polygon of external forces abedefa falling all in one straight line. 




Fig. 79. 



FUNICULAR POLYGON. 



TRIANGULAR TRUSS. 



147 



If, on the other hand, the whole load ae be given, we observe 
that this is borne by the stresses in the extreme bars AG and 
GE ; hence, drawing ag || AG, and eg || EG, we find eg and ga 
as the stresses in EG and GA respectively. Then, proceeding 
to the joint ABG, we find, since ga is the force exerted by 
GA at this point, that, drawing gb || GB, we shall have ab as 
the part of the load acting at the joint ABG, etc. 

§ 128. Funicular Polygon. — If the frame of Fig. 79 be 
inverted, we shall have the 
case of Fig. 80, where all 
the bars, except EG, are sub- 
jected to tension ; EG itself 
being subjected to compres- 
sion. The construction of the 
diagram of stresses being en- 
tirely similar to that already 
explained for Fig. 79, the ex- 
planation will not be repeated 
here. If the compression 
piece be omitted, the case 
becomes that of a chain hung 
at the upper joints (the supporting forces then becoming iden- 
tical with the tensions in the two extreme bars), the line gf 
would then be omitted from the diagram, and the polygon of 
external forces would become abcdega. 

§ 129. Triangular Truss : Wind Pressure. — Inasmuch as 
the pressure of the wind upon a truss is assumed to be normal 
to the rafter on which it blows, we will next consider the case 
of a triangular truss with the load distributed over one rafter 
only, and normal to the rafter. 




Fig. 80. 



There may be three cases : — 

i°. When there is a roller under one end, and the wind 
blows from the other side. 



148 



APPLIED MECHANICS. 



2°. When there is a roller under one end, and the wind 
blows from the side of the roller. 

3 . When there is no roller under either end. 

The last arrangement should always be avoided except in 
small and unimportant constructions ; for the presence of a 
roller under one end is necessary to allow the truss to change 
its length with the changes of temperature, and to prevent the 
stresses that would occur if it were confined. 



Case I. — Using Bow's notation, we have (Fig. 81) the 

whole load represented 
in the diagram by db. 
Its resultant acts at the 
middle of the rafter 
AE, whereas the sup- 
porting force at the 
right-hand end is (in 
consequence of the pres- 
ence of the roller) verti- 
cal. Hence, to find the 
line of action of the other 
supporting force, pro- 
duce the line of action 
of the load till it meets 
a vertical line drawn 
through the roller, and join their point of intersection with the 
support where there is no roller. We thus obtain CD as the 
line of action of the left-hand support. 

We can now determine the magnitude of the supporting- 
forces be and cd by constructing the triangle bedb of external 
forces. 

Now resolve the normal distributed force db into two single 
forces (equal to each other in this case), da and ab respectively, 
acting at the left-hand support and at the apex. 




Fig. 81. 



TRIANGULAR TRUSS: WIND PRESSURE. 



I49 



Now proceed to the left-hand support. We find four forces 
in equilibrium, of which two are entirely known ; viz., cd and 
da: hence, constructing the quadrilateral cdaec, we have ae as 
the thrust in AE, and ec as the tension in EC. 

Next proceed to the apex, and construct the triangle of 
equilibrium abea, and we obtain be as the thrust in BE. 

The triangle bceb is then the triangle of equilibrium for the 
right-hand sup- 
port. 

Case II. — 
In this case 
(Fig. 82) we fol- 
low the same 
method of pro- 
cedure, only the 
point of inter- 
section of the 
load and sup- 
porting forces 
is above, instead of below, the truss. The figure explains itself 

so fully that it is unnecessary to 

explain it here. 



Case III. — In this case the 
supports are capable of exerting 
resistance in any direction what- 
ever ; so that, if any circumstance 
should determine the direction 
of one of them, that of the other 
When there is no such circum- 




Fig. 82. 




Fig. 83. 



would be determined also, 
stance, it is customary to assume them parallel to the load 
( Fi g- 8 3)- Making this assumption, we begin, by dividing the 
line db, which represents the load, into two parts, inversely 



150 APPLIED MECHANICS. 

proportional to the two segments into which the line of action 
of the resultant of the load (the dotted line in the figure) 
divides the line EC. We thus obtain the supporting forces be 
and cd, and bedb is the triangle of external forces. We then 
follow the same method as in the preceding cases. 

§ 130. General Determination of the Stresses in Roof- 
Trusses. — In order to compute the stresses in the different 
members of a roof-truss, it is necessary first to know the 
amount and distribution of the load. 

This consists generally of — 

i°. The weight of the truss itself. 

2°. The weight of the purlins, jack-rafters, and superin- 
cumbent roofing, as the planks, slate, shingles, felt, etc. 

3 . The weight of the snow. 

4 . The weight of the ceiling of the room immediately 
below if this is hung from the truss, or the weight of the 
floor of the loft, and its load, if it be used as a room. 

5°. The pressure of the wind ; and this may blow from 
either side. 

6°. Any accidental load depending on the purposes for which 
the building is used. As an instance, we might have the case 
where a system of pulleys, by means of which heavy weights 
are lifted, is attached to the roof. 

In regard to the first two items, and the fourth, whenever 
the construction is of importance, the actual weights should 
be determined and used. In so doing, we can first make an 
approximate computation of the weight of the truss, and use it 
in the computation of the stresses ; the weights of the ceiling 
or of the floor below being accurately determined. After the 
stresses in the different members have been ascertained by the 
use of these loads ; and the necessary dimensions of the mem- 
bers determined, we should compute the actual weight of the 
truss ; and if our approximate value is sufficiently different 
from the true value to warrant it, we should compute again 



STRESSES IN ROOE-TRUSSES. 



151 



the stresses. This second computation will, however, seldom be 
necessary. 

In making these computations, the weights of a cubic foot 
of the materials used will be needed ; and average values are 
given in the following table with sufficient accuracy for the 
purpose. 



Weight of some Building Ma- 
terials per Cubic Foot. 



Pounds. 



Weight of Slating per Square 

Foot. 

According to Trautwine. 



Pounds. 



Timber. 

Chestnut 

Hemlock 

Maple 

Oak, live 

Oak, white 

Pine, white 

Pine, yellow, Southern 
Spruce 



Iron. 



Iron, cast . 
Iron, wrought 
Steel . . . 



Other Substances. 

Asphaltum 

Mortar, hardened . . 
Snow, freshly fallen 
Snow, compacted by rain 
Slate 



4i 
2 5 
4i 
59 
49 
25 to 30 

45 
2 z to ^o 



450 
480 
490 



80 to 90 
103 

5 t0 I2 

15 to 50 

140 to 180 



inch thick on laths 



_3_ 



i-inch boards 
li « - 
laths . . 
i-inch boards 
li « " 
laths . . 
i-inch boards 
t j, « « 



With slating-felt add 



Number of Nails in One Pound. 



3-penny 
4 



4-75 

6-75 

7-3° 

7.00 

9.00 

9-55 

9-25 

11.25 

11.80 

ilb. 

3 lbs. 



No. 



450 
340 
J50 
100 

60 
40 
25 



As to the weight of the snow upon the roof, Stoney recom- 
mends the use of 20 pounds per square foot in moderate 
climates ; and this would seem to the writer to be borne out by 
the experiments of Trautwine as recorded in his handbook, 



152 



APPLIED MECHANICS. 



although Trautwine himself considers 12 pounds per square 
foot as sufficient. 

§131. Wind Pressure. — The wind pressure, as has been 
stated, is not a vertical force, but is always assumed as acting 
normal to the roof on that side from which the wind blows. 
In order to determine the magnitude of this pressure, it has 
been advocated by some, to resolve the force of the wind into 
two components, one acting along, and one at right angles to, 
the roof. This method is, however, incorrect, because the 
moving air, striking against other particles of air, is deviated 
from its course, and the air, after striking the roof, cannot 
escape freely, but meets with the resistance of the surrounding 
air. Hence we must resort to experiment to determine the 
amount of the pressure due to the wind. The experiments of 
Hutton, according to Greene, give, for the pressure upon a 
surface whose inclination to the horizon is i, the value 



/'sin i 



. 1.84 cosz'— I 



where P is the pressure which would be exerted on the surface 
if normal to the wind ; and, taking the maximum force of the 
wind as 40 pounds per square foot, he gives for the normal 
pressures on surfaces inclined to the horizon the following : — 



Inclination. 


Normal Pressure. 


Inclination. 


Normal Pressure. 


5° 


5-2 


35° • 


30.I 


10° 


9.6 


40° 


334 


i5° 


I4.O 


45° 


36.1 


20° 


18.3 


5°° 


38.1 


*5° 


22.5 


55° 


39- 6 


3°° 


26.5 


6o° 


40 



and for all steeper angles 40. 



DISTRIBUTION OF THE LOADS. 1 53 

Later experiments show that these numbers are too small : 
stresses of 70 lbs. per square foot have been registered on vertical 
surfaces ; but we have no systematic set of tests on the subject. 

§132. Approximate Estimation of the Load. — In all 
important constructions, the estimates of the loads should be 
made as above described. For smaller constructions, and for 
the purposes of a preliminary computation in all cases, we only 
estimate the roof-weight roughly ; and some authors even as- 
sume the wind pressure as a vertical force. 

Trautwine recommends the use of the following figures for 
the total load per square foot, including wind and snow, when 
the span is 75 feet or less : — 

Roof covered with corrugated iron, unboarded ... 28 lbs. 

Roof plastered below the rafters 38 " 

Roof, corrugated iron on boards 31 " 

Roof plastered below the rafters 41 " 

Roof, slate, unboarded or on laths 33 " 

Roof, slate, on boards ij inches thick 35 " 

Roof, slate, if plastered below the rafters 46 " 

Roof, shingles on laths 30 " 

Roof plastered below rafters or below tie-beam ... 40 " 
From 75 to 100 feet, add 4 lbs. to each. 

§ 133. Distribution of the Loads. — The methods for de- 
termining the stresses, which will be used here, give correct 
results only when the loads are concentrated at joints, and are 
not distributed over any members of the truss. 

In constructions of importance, this concentration of the 
loads at the joints should always be effected if possible ; 
because, when this is the case, the stresses in the members 
of the truss can be, if properly fitted, obliged to resist only 
stresses of direct tension, or of direct compression ; but, when 
there is a load distributed over any member of the truss, that 
member, in addition to the direct compression or direct tension, 
is subjected to a bending-stress. The effect of this bending 



154 



APPLIED MECHANICS. 



cannot be discussed until we have studied the theory of beams. 
Nevertheless, it is a fact that we have no experimental knowl- 
edge of the behavior of a piece under combined compression 
and bending ; and we know that when certain pieces are to 
resist bending, in addition to tension, they must be made much 
larger in proportion than is necessary when they bear tension 



only. 




Fig. 84. 

The manner in which this concentration of the loads is 
effected, is shown in Fig. 84, which is intended to represent one 
of a series of trusses that supports a roof, the rafters being the 
two lower ones in the figure. Resting on two consecutive 
trusses, and extending from one to the other, are beams called 
purlins, which should be placed only above the joints of the truss, 
and which are shown in cross-section in the figure. On these 
purlins are supported the jack-rafters parallel to the rafters, and 
at sufficiently frequent intervals to support suitably the plank 
and superincumbent roofing-materials. 

By this means we secure that the entire bending-stress comes 
upon the jack-rafters and purlins, and that the rafters proper 
are subjected only to a direct compression. When, however, 
the load is distributed, i.e., when the roofing rests directly on the 
rafters, or when the purlins are placed at points other than the 
joints, the bending-stress should be taken into account ; and 
while the methods to be developed here will give the stresses 



DIRECT DETERMINATION OF THE STRESSES. 1 55 

in all the members that are not subjected to bending, the bend- 
ing-stress may be largely in excess of the direct stress in those 
pieces that are subjected to bending. How to take this into 
account will be explained later. 

Another important item to consider is, that, in constructions 
of importance, a roller should be placed under one end of the 
truss to allow it to move with the change of temperature ; as 
otherwise some of the members will be either bent, or at least 
subjected to initial stresses. The presence of a roller obliges 
the supporting force at that point to be vertical, whether the 
load be vertical or inclined. 

It is customary, and does not entail any appreciable error, 
to consider the weight of the truss itself, as well as that of the 
superincumbent load, as concentrated at the upper joints ; i.e., 
those on the rafters. 

In the case of a ceiling on the room below, or of a loft 
whose floor rests on the lower joints, we must, of course, ac- 
count the proper load as resting on the lower joints. 

§134. Direct Determination of the Stresses. — This, as 
we have seen, is merely a question of equilibrium of forces in 
a plane, where certain forces acting are known, and others are 
to be determined. 

As to the methods of solution, we might adopt — 

i°. A graphical solution, laying off the loads to scale, and 
constructing the diagram by the use of the propositions of 
the polygon, and the triangle of forces, and then determining the 
results by measuring the lines representing the stresses to 
the same scale. 

2°. An analytical solution, imposing the analytical conditions 
of equilibrium, as given under the " Composition of Forces," 
between the known and unknown forces. 

3°. A third method is to construct the diagram as in the 
graphical solution, but then, instead of determining the results 
by measuring the resulting lines to scale, to compute the un- 



156 APPLIED MECHANICS. 

known from the known lines of the diagram by the ordinary 
methods of trigonometry. 

The first, or purely graphical, method, is one which has 
received a very large amount of attention of late years, and 
in which a great deal of progress has been made. It is, doubt- 
less, very convenient for a skilled draughtsman, and especially 
convenient for one who, though skilled in draughting, is not 
free with trigonometric work ; but it seems to me, that, when 
the results are determined by computation from the diagram, 
there is less chance of a slight error in some unfavorable tri- 
angle vitiating all the results. I am therefore disposed to 
recommend for roof-trusses the third method. 

In the case of bridge-trusses, on the other hand, I believe 
the graphical not to be as convenient as a purely analytic 
method. 

§ 135. Roof-Trusses. — In what follows, the graphical solu- 
tions will be explained with very little reference to the trigono- 
metric work, as that varies in each special case, and to one who 
has a reasonable familiarity with the solution of plane triangles, 
it will present no difficulty ; whereas to deduce the formulae 
for each case would complicate matters very much. Proceed- 
ing to special examples, let us take, first, the truss shown in 
Fig. 85, and let the vertical load upon it be W uniformly dis- 
tributed over the top of the roof, the purlins being at the joints 
on the rafters. 

The loads at the several joints will then be as follows, viz. 
(Fig. 85a), — 

ab = kl = — , be = ed = de = ef = fg = gh = hk = ^-. 
16 o 



Then the supporting forces will be 

lm = ma = 
We thus have, as polygon of external forces, abcdcfghklma. 



/ w 

lm = ma = — . 

2 



ROOF-TRUSSES. 



157 



Now proceed to either support, say, the left-hand one 
there we have the two forces ab and ma known, while b 
ym are unknown. We then construct 
the quadrilateral maby in the figure, and 
thus determine by and ym. As to whether 



; and 
v and 





Fig. 85. 

these represent thrust or tension, 

we need only remember that they 

are the forces exerted by the re- FlG ' B5C ' 

spective bars at the joints: and, since by is directed away from 

the bar BY, this bar is in compression; whereas, ym being 

directed towards the bar YM, that bar is in tension. 



158 APPLIED MECHANICS. 

Having determined these two stresses, we next proceed to 
another joint, where we have only two unknown forces. Take 
the joint at which the load be acts, and we have as known 
quantities the load be, and also the force exerted by the bar 
YB y which is in compression. This force is now directed away 
from the bar, and hence is represented by yb. The unknown 
forces are the stresses in CX and XY. Hence we construct 
the quadrilateral cxybc ; and we thus determine the stresses in 
CX and XY as ex and xy, both being thrusts. 

Next proceed to the joint YXIV, and construct the quadri- 
lateral myxwm, and thus determine the tension xw and the 
tension wm. 

Next proceed to the joint where cd acts, and so on. We 
thus obtain the diagram (Fig. 85^) giving all the stresses. 

The truss in the figure was constructed with an angle of 30 
at the base, and hence gives special values in accordance with 
that angle. 

In order to show how we may compute the stresses from the 
diagram, the computation will be given. 

From triangle bmy, we have bm = ~ W 

.'. ym = 1-Wcot 30 = ^W 
16 16 

by — —W cosec 30 = - W = ky. 
16 8 

From the triangle time, we have cm = ^- IV, 

16 



um = 5V3 jp; 

16 ' 



1 
yw = - yu 
2 



;fe *V5) - & 



ROOF-TRUSSES. 



159 



yx — yw sec -10° = ( ^ W\^- = — = xv = vt 
\i6 fa 8 



, W W 

xw = \yx = — , #p = — , 



j* 



= <^) = 



^, 



k/» 



y V 2^6 2^6 ' 16 



V V 256 256 8 

\ 8 )\fi 4 



^je = o/zh sec 30" = ( ^j-^ W)-^=. — -W, 



vd = um sec 3 



o = /5VS^\ » = 5^ 

V16 ;v 3 8 

*/ = f^L3LJv\4= = -W. 

V 4 / N/ 3 2 



Hence we shall have for the stresses, 



Rafters 


(compression) . 


Verticals 


(tension). 


£# = ho 






xw = op 


._ w 

16' 


dv = gq 
ct =fs 




= \w. 


vn = qr 
ts = 


_ w 

8* 
8 


Horizontal Ties 


(tension). 










= 1^3^ 

16 


Diagonal Braces (compression). 


my = mn 




xy = ou 


8 ' 


mw = mp 




8 


wv = qp 


J&w. 

16 


mu = mr 




16 


tu = sr 


8 



l60 APPLIED MECHANICS. 

Next, as to the stresses due to wind pressure, we will sup- 
pose that there is a roller under the left-hand end of the truss, 
and none under the right-hand end ; and we will proceed to 
determine the stresses due to wind pressure. 

First, suppose the wind to blow from the left-hand side of 
the truss, and let the total wind pressure be (Fig. 85$) af '= W x . 
The resultant, of course, acts along the dotted line drawn per- 
pendicular to the left-hand rafter at its middle point, as shown 
in Fig. 85. 

The left-hand supporting force will be vertical : hence, pro- 
ducing the above-described dotted line, and a vertical through 
the roller to their intersection, and joining this point with the 
right-hand end of the truss, we have the direction of the right- 
hand supporting force. In this case, since the angle of the 
truss is 30 , the line of action of the right-hand supporting 
force coincides in direction with the right-hand rafter. We 
now construct the triangle of external forces afm, and we 
obtain the supporting forces fm and ma. We then have, as 
the loads at the joints, 

ab = -j± = ef, 

be e= — - = cd = de. 



Then proceed as before to the left-hand joint ; and we find that 
two of the four forces acting there are known, viz., ma and ab, 
and two are unknown, viz., the stresses in BY and YM. Then 
construct the quadrilateral mabym, and we have the stresses by 
and ym ; the first being compression and the second tension, 
as shown by reasoning similar to that previously adopted. 

Then pass to the next joint on the rafter, and construct the 
quadrilateral ybcxy, where yb and be are already known, and we 
obtain ex and xy ; and so proceed as before from joint to joint, 



ROOF-TRUSSES WITH LOADS AT LOWER JOINTS. l6l 



remembering, that, in order to be able to construct the polygon 
of forces in each case, it is necessary that only two of the forces 
acting should be unknown. 

When the wind blows from the other side, we shall obtain 
the diagram shown in Fig. 8$c. 

After having determined the stresses from the vertical load 
diagram and those from the two wind diagrams, we should, in 
order to obtain the greatest stress that can come on any one 
member of the .truss, add to the stress due to the vertical load 
the greater of the stresses due to the wind pressure. 

§136. Roof-Truss with Loads at Lower Joints. — In 
Fis:. 86 is drawn a stress diagram 
for the truss shown in Fig. 84 on 
the supposition that there is also 
a load on the lower joints. In 
this case let W be the whole load 
of the truss, except the ceiling, 
and W 1 the weight of the ceiling 
below ; the latter being supported 
at the lower joints and on the 
two extreme vertical suspension 
rods. Then will the loads at the joints be as follows ; viz., — 




Fig. 86. 



ab = ^(W+ JV Z ) 

be = \(W + tV z ) 

cd = \W 

mn = \W X = rq 



= hk, 

= gh = de = fg = ef, 

= on = qp = op. 



Observe that there is no joint at the lower end of either of the 
end suspension rods, but that whatever load is supported by 
these is hung directly from the upper joints, where be and hk act. 
We have also for each of the supporting forces Im and ra 



\{W+ W x ). 



1 62 APPLIED MECHANICS. 

Hence we have, for the polygon of external forces, 

abcdefghklm nopqra, 

which is all in one straight line, and which laps over on 
itself. 

In constructing the diagram, we then proceed in the same 
way as heretofore. 

§ 137. General Remarks. — As to the course to be pursued 
in general, we may lay down the following directions : — 

I °. Determine all the external forces ; in otJier words, the loads 
being known, determine the supporting forces. 

2°. Construct the polygon of forces for each joint of the truss, 
beginning at some joint where only two of the forces acting at 
that joint are unknown. This is usually the case at the support. 
Then proceed from joint to joint, bearing in mind that we can 
only determine the polygon of forces when the magnitudes of 
all but two sides are known. 

3 . Adopt a certain direction of rotation, and adhere to it 
throughout ; i.e., if we proceed in right-handed rotation at one 
joint, we must do the same at all, and we shall thus obtain neat 
and convenient figures. 

4 . Observe that the stresses obtained are the forces exerted 
by the bars under consideration, and that these are thrusts when 
they act away from the bars, and tensions when they are directed 
towards the bars. 

We will next take some examples of roof-trusses, and con- 
struct the diagrams of some of them, calling attention only to 
special peculiarities in those cases where the)' exist. 

It will be assumed that the student can make the trigono- 
metric computations from the diagram. 

The scale of load and wind diagram will not always be the 
same ; and the stress diagrams will in general be smaller than 
is advisable in using them, and very much too small if the 



ROOF-TRUSSES WITH LOADS AT LOWER JOINTS. 1 63 

results were to be obtained by a purely graphical process with- 
out any computation. 

The loads will in all cases be assumed to be distributed 
uniformly over the jack-rafters, or, in other words, concen- 
trated at the joints. 

Those cases where no stress diagram is drawn may be con- 
sidered as problems to be solved. 




Fig. 87. 



Fig. 87*. 




Fig. 873. 



Fig. 8 7 c 



Fig. 87^. 



164 



APPLIED MECHANICS. 




Fig. 88. 





\fghkl 




abode 



Fig. 88<r. 



ROOF-TRUSSES WITH LOADS AT LOWER JOINTS. l6$ 




Fig. 89. 



Fig. Bga. 




Fig. 90. 



Fig. goa. 



Fig. 91. 




c A 





C>sp 






/M \ 


\E 


B/^ 




/L 


O 


N\ 


K 


H 


G 



Fig. 92. 



Fig. Q2rt. 



Fig. 93. 



Fig. 93a. 



1 66 



APPLIED MECHANICS. 



§138. Hammer-Beam Truss (Fig. 94). — This form of 
truss is frequently used in constructions where architectural 
effect is the principal consideration rather than strength. It 
is not an advantageous form from the point of view of strength, 





Fig. 94. 



Fig. 94a. 





Fig. 943. 



Fig. 



for the absence of a tie-rod joining the two lower joints causes 
a tendency to spread out at the base, which tendency is usually 
counteracted by the horizontal thrust furnished by the bat- 
tresses against which it is supported. 



HAMMER-BEAM TRUSS. 1 67 

When such a thrust is furnished (or were there a tie-rod), 
and the load is symmetrical and vertical, the bars are not all 
needed, and some of them are left without any stress. In 
the case in hand, it will be found that UV, VM, MQ, and QR 
are not needed. We must also observe that the effect of the 
curved members MY, MV, MQ, and MN 'on the other parts of 
the truss is just the same as though they were straight, as 
shown in the dotted lines. The curved piece, of course, has to 
be -subjected to a bending-stress in order to resist the stress 
acting upon it. If, as is generally the case, the abutments are 
capable of furnishing all the horizontal thrust needed, it will 
first be necessary to ascertain how much they will be called 
upon to furnish. To do this, observe that we have really a truss 
similar to that shown in Fig. 92, supported on two inclined 
framed struts, of which the lines of resistance are the dotted 
lines (Fig. 94) 1 4 and 7 8, and that, under a symmetrical load, 
this polygonal frame will be in equilibrium, and, moreover, the 
curved pieces MV and MQ will be without stress, these only 
being of use to resist unsymmetrical loads, as the snow or 
wind. 

Let the whole load, concentrated by means of the purlins 
at the joints of the rafters, be IV. Then will the truss 46 7 have 

to bear J W> and this will give — to be supported at each of 

4 
the points 4 and 7. Moreover, on the space 2 4 is distributed 

W 

— , which has, as far as overturning the strut is concerned, the 

4 

same effect as — at 2, and — at 4. Hence the load to be sup- 
8 8 

ported at 4 by the inclined strut is a vertical load equal to 

(i + i) W — \ W. We may then find the force that must be 

furnished by the abutment, or by the tie-rod, in either of the 

two following ways : — 



1 68 APPLIED MECHANICS. 

i°. By constructing the triangle ySe (Fig. 94*2), with yS = 
I IV, ye II 14, and cS parallel to the horizontal thrust of the abut- 
ment ; then will ySe be the triangle of forces at 1, and eS will be 
the thrust at 1. 

2°. Multiply § W by the perpendicular distance from 4 to 
1 2, and divide by the height of 4 above 1 8 for the thrust of the 
abutment ; in other words, take moments about the point 1. 

Now, to construct the diagram of stresses, let, in Fig. 94^, 
the loads be 

ab, be, cd, de, ef, fg, gh, hk, and kl, 
and let 

lz = za = \W 

be the vertical component of the supporting force ; let sm be 
the thrust of the abutment : then will Im and ma be the real 
supporting forces ; and we shall have, for polygon of external 
forces, 

abcdefghklma. 

Then, proceeding to the joint 1, we obtain, for polygon of forces, 

maym ; 

and, proceeding from joint to joint, we obtain the stresses in all 
the members of the truss, as shown in Fig. 94$. 

It will be noticed that UV and RQ are also free from 
stress. 

If we had no horizontal thrust from the abutment, and the 
supporting forces were vertical, the members MV and MQ 
would be called into action, and M Fand MN would be inactive. 
To exhibit this case, I have drawn diagram 94^, which shows 
the stresses that would then be developed. A Fand NL would 
become merely part of the supports. 

In this latter case the stresses are generally much greater 
than in the former, and a stress is developed in UV. 



SC1SS0R-BEAM TRUSS. 



169 



§ 139. Hammer-Beam Truss: Wind Pressure. — Fig. 95 
shows the stress diagram of the hammer-beam truss for wind 
pressure when there is no roller under either end, and when 
the wind blows from the left. A similar diagram would give the 
stresses when it blows from the right. 




Fig. 95. 



Fig. 950. 



The cases when there is a roller are not drawn : the student 
may construct them for himself. 

§140. Scissor-Beam Truss. — We have already discussed 
two forms of scissor-beam truss 
in Figs. 90 and 91. These 
trusses having the right number 
of parts, their diagrams present 
no difficulty. Another form of 
the scissor-beam truss is shown 
in Fig. 96, and its diagram pre- 
sents no difficulty. 

The only peculiarity to be noticed is, that, after having con- 
structed the polygon of external forces, 

abcdefma, 

we cannot proceed to construct the polygon of equilibrium for 
one of the supports, because there are three unknown forces 




Fig. 96. 



Fig. 96a. 



170 



APPLIED MECHANICS. 



there. We therefore begin at the apex CD, and construct the 
triangle of forces cdl for this point ; then proceed to joint CB, 
and construct the quadrilateral 

bclkb ; 

then proceed to the left-hand support, and obtain 

mabkgm ; 
and so continue. 

§ 141. Scissor-Beam Truss without Horizontal Tie. — 
Very often the scissor-beam truss is constructed without any 
horizontal tie, in which case it has the appearance of Fig. 97, 
where there is sometimes a pin at GKLH and sometimes not. 





Fig. 97a. 




Fig. 97. 



Fig. 973. 



X A 

1 



Fig. 97c. 



In this case, if the abutments are capable of furnishing hori- 
zontal thrust to take the place of the horizontal tie of Fig. 96, 
we are reduced back to that case. If the abutments are not 
capable of furnishing horizontal thrust, we are then relying on 
the stiffness of the rafters to prevent the deformation of the 
truss ; for, were the points BC and DE really joints, with pins, 
the deformation would take place, as shown in Fig. 97^ or Fig. 
97^, according as the two inclined ties were each made in one 
piece or in two (i.e., according as they are not pinned together 
at KH, or as they are pinned). This necessity of depending 
on the stiffness of the rafters, and the liability to deformation 
if they had joints at their middle points, become apparent as 
soon as we attempt to draw the diagram. Such an attempt is 






SCISSOR-BEAM TRUSS WITHOUT HORIZONTAL TIE. 171 

made in Fig. 97c, where abcdefga is the polygon of external 
forces, gabkg the polygon of stresses for the left-hand support, 
kbclk that for joint BC. Then, on proceeding to draw the tri- 
angle of stresses for the vertex, we find that the line joining d 
and / is not parallel to DL y and hence that the truss is not 
stable. We ought, however, in this latter case, when the sup- 
porting forces are vertical, and when we rely upon the stiffness 
of the rafters to prevent deformation, to be able to determine 
the direct stresses in the bars ; and for this we will employ an 
analytical instead of a graphical method, as being the most con- 
venient in this case. 

Let us assume that there is no pin at the intersection of the 
two ties, and that the two rafters are inclined at an angle of 45 ° 
to the horizon. 

We then have, if W = the entire load, and a == angle 
between BK and KG, 

ab = ef = — , be = ed = de = — , 
J 8' 4 

, . 1 2 

tana = j;, sin a = — , cos a = — , 

Let x be the stress in each tie, and let y = cl = dl = thrust 
in each upper half of the rafters. 

Then we must observe that the rafter has, in addition to its 
direct stresses, a tendency to bend, due to a normal load at the 
middle, this normal load being equal to the sum of the normal 
components of be and of x, when these are resolved along and 
normal to the rafter. Hence 

normal load = ^cosa -\ sin 4 c . 

4 

This, resolved into components acting at each end of the rafter, 
gives a normal downward force at each end equal to 

-|vecosa -f- J^sin45°. 



172 APPLIED MECHANICS. 

Hence, resolving all the forces acting at the left-hand support 
into components along and at right angles to the rafter, and 
imposing the condition of equilibrium that the algebraic sum 
of their normal components shall equal zero, we have, if we call 
upward forces positive, 

-§ W^sin45° — (J#cosa -+- •JW 7 sin45°) — #sina = o; (i) 

but, since 

#cosa = 2X sin a, 

we have from (i) 

W . 
2x sin a = — sin 4s 

W . 
.'. #sina == — sm45 
8 

... x = i W sjMf. (2) 

8 sin a v ' 

Then, proceeding to the apex of the roof, we have that the load 

4 
gives, when resolved along the two rafters, a stress in each 

equal to 

W . 

— sin 45 . 

4 
Hence the load to be supported in a direction normal to the 
rafter at the apex is 

— sin 45 + (i^cosa + —sin 45 ). 
4 k 

Hence, substituting for x its value, we have 

y = d=dl= ^sin 45 °. (3) 

2 

Then, proceeding to the left-hand support, and equating to zero 
the algebraic sum of the components along the rafter, we have 

bk = (ga — #/;)cos45° + ^ COSa 

= f ^sin 45° + 1 ^sin 45° = § ^sin 45°- (4) 



SCISSOR-BEAM TRUSS WITHOUT HORIZONTAL TIE. 1 73 

We have thus determined in (2), (3), and (4) the values of x, y, 
and bk = eh. 

By way of verification, proceed to the middle of the left- 
hand rafter, and we find the algebraic sum of the components 
of be and x along the rafter to be 

JJFcos45° — ^sina =-J^sin45°; 

and this is the difference between bk and cl> as it should be. 

We have thus obtained the direct stresses ; and we have, in 
addition, that the rafter itself is also subjected to a bending- 
moment from a normal load at the centre, this load being equal 
to 

#cosa H sin 45 = — sin 45 . 

4 2 

How to take this into account will be explained under the 
" Theory of Beams." 

§142. Examples. — The following figures of roof-trusses 
may be considered as a set of examples, for which the stress 
diagrams are to be worked out. 

Observe, that, wherever there is a joint, the truss is to be. 
supposed perfectly flexible, i.e., free to turn around a pin. 




Fig. 98. 



Fig. 99. 



zf^^ ^^ 



Fig. 100. 



Fig. i 01. 




Fig. 102. 




Fig. 103. 



Fig. 104. 




Fig. 105. 




Fig. 106. 




Fig. 



Fig. 108 



74 APPLIED MECHANICS. 



CHAPTER IV. 
BRIDGE- TR USSES. 

§ 143. Method of Sections. — It is perfectly possible to 
determine the stresses in the members of a bridge-truss 
graphically, or by any methods that are used for roof-trusses. 

In this work an analytical method will be used ; i.e., a method 
of sections. This method involves the use of the analytical con- 
ditions of equilibrium for forces in a plane explained in § 63. 
These are as follows ; viz., — 

If a set of forces in a plane, which are in equilibrium, be 
resolved into components in two directions at right angles to 
each other, then — 

i°. The algebraic sum of the components in one of these 
directions must be zero. 

2 . The algebraic sum of the components in the other of 
these directions must be zero. 

3 . The algebraic sum of the moments of the forces about 
any axis perpendicular to the plane of the forces must be zero. 

Assume, now, a bridge-truss (Figs. 109, no, III, 112, pages 
176 and 177) loaded at a part or all of the joints. Conceive a 
vertical section ab cutting the horizontal members 6-8 and 7-9 
and the diagonal 7-8, and dividing the truss into two parts. 
Then the forces acting on either part must be in equilibrium, 
in other words, the external forces, loads, and supporting forces, 
acting on one part, must be balanced by the stresses in the 
members cut by the section ; i.e., by the forces exerted by the 
other part of the truss on the part under consideration. Hence 
we must have the three following conditions; viz., — 



SHEARING-FORCE AND B ENDING-MOMENT. 1 75 

i°. The algebraic sum of the vertical components of the 
above-mentioned forces must be zero. 

2°. The algebraic sum of the horizontal components of these 
forces must be zero. 

3 . The algebraic sum of the moments of these forces about 
any axis perpendicular to the plane of the truss must be zero. 

§ 144. Shearing-Force and Bending-Moment. — Assum- 
ing all the loads and supporting forces to be vertical, we shall 
have the following as definitions. 

The Shearing-Force at any section is the force with which 
the part of the girder on one side of the section tends to slide 
by the part on the other side. 

In a girder free at one end, it is equal to the sum of the 
loads between the section and the free end. 

In a girder supported at both ends, it is equal in magnitude 
to the difference between the supporting force at either end, 
and the sum of the loads between the section and that support- 
ing force. 

The Bending-Moment at any section is the resultant moment 
of the external forces acting on the part of the girder to one side 
of the section, tending to rotate that part of the girder around 
a horizontal axis lying in the plane of the section. 

In a girder free at one end, it is equal to the sum of the 
moments of the loads between the section and the free end, 
about a horizontal axis in the section. 

In a girder supported at both ends, it is the difference be- 
tween the moment of either supporting force, and the sum of 
the moments of the loads between the section and that sup- 
port ; all the moments being taken about a horizontal axis in 
the section. 

§ 145. Use of Shearing-Force and Bending-Moment. — 
The three conditions stated in § 143 may be expressed as fol- 
lows : — 

i°. The algebraic sum of the horizontal components of the 
stresses in the members cut by the section must be zero. 



7 6 



APPLIED MECHAXICS. 



2°. The algebraic sum of the vertical components of the 
stresses in the members cut by the section must balance the 
shearing-force. 

3°. The algebraic sum of the moments of the stresses in 
the members cut by the section, about any axis perpendicular to 
the plane of the truss, and lying in the plane of the section, 
must balance the bending-moment at the section. 

As the conditions of equilibrium are three in number, they 
will enable us to determine the stresses in the members, pro- 
vided the section does not cut more than three ; and this 
determination will require the solution of three simultaneous 
equations of the first degree with three unknown quantities 
(the stresses in the three members). 

By a little care, however, in choosing the section, we can 
very much simplify the operations, and reduce our work to the 
solution of one equation with only one unknown quantity ; the 
proper choice of the section taking the place of the elimination. 

§146. Examples of Bridge-Trusses. — Figs. 109-1 12 rep- 
resent two common kinds of bridge-trusses : in the first two 

the braces are all 



i\5S2 



6 b 



a 9 



x 11 13 15 



19 21 23 25 27 29 



7 W\AAAAA/\A/ ^ 

8 2 10 12 14 16 18 20 22 24 26 28 ^ 



diagonal, in the 

last two they are 

partly vertical and 

FlG - I09 - partly diagonal. 

The first two are called Warren girders, or half-lattice girders ; 

since there is only one system of bracing, 






as in the figures. When, on the other 

i A f 3 5 7[g 9Kr 11 l i 

hand, there are more than one system, so j ^\/\j\{\ J \J\l 
that the diagonals cross each other, they | 
are called lattice girders. 



10 12 



47. General Outline of the Steps 



Fig. ho. 



to be taken in determining the Stresses 
in a Bridge-Truss under a Fixed Load. 

i°. If the truss is supported at both ends, find the sup- 
porting forces. 



DETERMINING THE STRESSES IN A BRIDGE-TRUSS. IJJ 





l 3 


5 la 


9 


tl 13 15 17 19 21 23 25 27 23 


-1 — 


V 


\ 


\ 


\ 




\ 


X 


XIX 


X 


X 


X 


X 


X 


/ 






\ 














8 








1 1 


i 4 60 


10 12 1* 16 18 20 22 £± «o 


1 


J -r 


























11 13 



Fig. ii2. 



2°. Assume, in all cases, a section, in such a manner as not 
to cut more than three members if possible, or, rather, three 
of those that 
are brought 
into action 
by the loads 
on the truss ; 

and it will FlG - «*■ 

save labor if we assume the section so as to cut two of the 

three very near their point of inter- 
section. 

3°. Find the shearing-force at the 
section. 

4°. Find the bending-moment at 
the section. 

5°. Impose the analytical conditions of equilibrium on all 
the forces acting on the part of the girder to one side of the 
section, — the part between the section and the free end when 
the girder is free at one end, or either part when it is supported 
at both ends. 

In the cases shown in Figs. 109 and no, we may describe 
the process as follows ; viz., — 

(a) Find the stress in the diagonal from the fact, that (since 
the stress in the diagonal is the only one that has a vertical 
component at the section) the vertical component of the stress 
in the diagonal must balance the shearing-force. 

(b) Take moments about the point of intersection of the 
diagonal and horizontal chord near which the section is taken ; 
then the stresses in those members will have no moment, so 
that the moment of the stress in the other horizontal must 
balance the bending-moment at the section. Hence the stress 
in the horizontal will be found by dividing the bending-moment 
at the section by the height of the girder. 

The above will be best illustrated by some examples. 



1/8 APPLIED MECHANICS. 

Example I. — Given the semi-girder shown in Fig. no, 
loaded at joint 13 with 4000 pounds, and at each of the joints 
!> 3> 5> 7> 9» an d 11 with 8000 pounds. Suppose the length of 
each chord and each diagonal to be 5 feet. Required the stress 
in each member. 

Solution. — For the purpose of explaining the method of 
procedure, we will suppose that we desire to find first the 
stresses in 8-10 and 9-10. 

Assume a vertical section very near the joint 9, but to the 
right of it, so that it shall cut both 8-10 and 9-10. 

If, now, the truss were actually separated into two parts at 
this section, the right-hand part would, in consequence of the 
loads acting on it, separate from the other part. This tendency 
to separate is counteracted by the following three forces : — 

i°. The pull exerted by the part g-x of the bar 9- 11 on the 
part ;r-n of the same bar. 

• 2°. The thrust exerted by the part 8-# of the bar 8-10 on 
the part ^-10 of the same bar. 

3 . The pull exerted by the part g-y of the bar 9-10 on the 
part y—\o of the same bar. 

The shearing-force at this section is 

8000 + 4000 = 12000 lbs., 

and this is equal to the vertical component of the stress in the 
diagonal. Hence 

Stress in 9-10 = — - = 12000(1.1547) = 13856 lbs. 

sin 60 

This stress is a pull, as may be seen from the fact, that, in 
order to prevent the part of the girder to the right of the 
section from sliding downwards under the action of the load, 
the part g-y of the diagonal 9-10 must pull the part y-10 of 
the same diagonal. 

Next take moments about 9: and, since the moment of the 
stresses in 9-1 1 and 9-10 about 9 is zero, we must have that the 
moment of the stress in 8-10; i.e., the product of this stress 
by the height of the girder, must equal the bending-moment. 



DETERMINING THE STRESSES IN A BRIDGE-TRUSS. 1 79 



Hence 



The bending-moment about 9 is 
8000 x 5 + 4000 x 10 



80000 foot-lbs. 



Stress in 8-10 = = 80000(0.23094) = 18475 lbs. 



4-33 



Proceed in a similar way for all the other members. The 
work may be arranged as in the following table ; the diagonal 
stresses being deduced from the shearing-forces by multiplying 
by 1. 1547, and the chord stresses from the bending-moments 
by multiplying by 0.23094. 



c 




Stresses in Diagonals cut 




Stresses in Chords opposite the 


e .£» 
.2 E 

in 


Shearing- 
Force 
in lbs. 


by Section, in lbs. 


Bending- 

Moment, in 

foot-lbs. 


respective Joints. 


Tension. 


Compression. 


Tension. 


Compression. 


I 


44OOO 


50806 




72OOOO 




166277 


2 


44OOO 




50806 


6lOOOO 


I40873 




3 


360OO 


4I5 6 9 




5OO00O 




II5470 


4 


360OO 




41569 


4IOOOO 


94685 




5 


28000 


3 2 33! 




32OOOO 




739OI 


6 


28000 




3 2 33* 


25OOOO 


57735 




7 


20000 


23094 




180OOO 




41569 


8 


20000 




23094 


I3OOOO 


30022 




9 


12000 


13856 




8OO0O 




18475 


10 


12000 




13856 


5OOOO 


"547 




11 


4OOO 


4619 




20000 




4618 


12 


4OOO 




4619 


I OOOO 


2309 





Example II. — Given the truss (Fig. 109) loaded at each of 
the lower joints with 10000 lbs. : find the stresses in the members. 
The length of chord is equal to the length of diagonal = 10 ft. 

In future, tensions will be written with the minus, and com- 
pressions with the plus sign. 

Solution. — Total load == 14(10000) = 140000 lbs. 
Each supporting force = 70000 " 

The entire work is shown in the following: tables : — 



i8o 



APPLIED MECHANICS. 









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o 
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m 


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DETERMINING THE STRESSES IN A BRIDGE-TRUSS. l8l 



Numbers of Diagonals. 


Stresses 


in Diagonals, in lbs. 


I- 2 


28-29 


— 70000 x 


1. 1547 = —80829 


2- 3 


27-28 


+ 60000 x 


I.T547 = +69282 


3" 4 


26-27 


— 60000 X 


1. 1547 = —69282 


4- 5 


25-26 


+ 50000 x 


1. 1547 = +57735 


5-6 


24-25 


— 50000 x 


1. 1547 = -57735 


6- 7 


23-24 


+ 40000 x 


1. 1547 = +46188 


7" 8 


22—23 


— 40000 x 


1. 1547 = —46188 


8-9 


21-22 


+ 30000 x 


I.I547 = +34641 


9-10 


20-21 


— 30000 x 


I - I 547 = -34641 


10-11 


19-20 


+ 20000 X 


i-i547 = +23094 


11-12 


18-19 


— 20000 x 


1. 1547 = -23094 


. 12 - J 3 


17-18 


+ 10000 x 


1. 1547 = +11547 


13-M 


16-17 


— 1 0000 x 


1. 1547 = -11547 


14-15 


15-16 


+0 






UPPER CHORDS. 



Numbers of Chords. 


Stresses in Chords, in lbs. 


"" 3 


27-29 


650000 X 0.11547 == + 75056 


3- 5 


25-27 


1200000 X 0.11547 = +138564 


5" 7 


23-25 


1650000 X 0.11547 = +190526 


7- 9 


21-23 


2000000 X 0.1 1547 = +230940 


9-1 1 


19-21 


2250000 X 0.11547 = +259808 


11-13 


17-19 


2450000 X 0.11547 = +282902 


r 3~i5 


15-17 


2450000 X 0.11547 = +282902 



182 



APPLIED MECHANICS. 



LOWER CHORDS. 



Numbers of Chords. 


Stresses in Chords, in lbs. 


2- 4 


26-28 


- 35 0000 X 0.11547 = - 40415 


4- 6 


24-26 


— 950000 X 0.1 1547 = —109697 


6- 8 


2 2-24 


— 1450000 X 0.11547 = —167432 


8-io 


20-2 2 


— 1850000 X 0.11547 = —213620 


IO-I2 


I8-20 


— 2150000 X 0.11547 = —248261 


12-14 


l6-l8 


-2350000 X 0.11547 = -267355 


I4-16 




— 2450000 X 0.11547 = —282902 



Example III. — Given the same truss as in Example II., 
loaded at 2, 4, 6, 8, 10, and 12 with 10000 lbs. at each point, 
the remaining lower joints being loaded with 50000 lbs. at each 
joint : find the stresses in the members. 

Example IV. — Given a semi-girder, free at one end (Fig. 
112), loaded at 2, 4, and 6 with 10000 lbs., and at 8, 10, and 12 
with 5000 lbs. : find the stresses in the members. 



TRAVELLING-LOAD. 

§148. Half-Lattice Girder: Travelling-Load. — When a 
girder is used for a bridge, it is not subjected all the time to 
the same set of loads. 

The load in this case consists of two parts, — one, the dead 
load, including the bridge weight, together with any permanent 
load that may rest upon the bridge ; and the other, the moving 
or variable load, also called the travelling-load, such as the 
weight of the whole or part of a railroad train if it is a railroad 
bridge, or the weight of the passing teams, etc., if it is a common- 
road bridge. Hence it is necessary that we should be able to 
determine the amount and distribution of the loads upon the 
bridge which will produce the greatest tension or the greatest 



GREATEST DIAGONAL STRESSES IN GIRDER. 1 83 

compression in every member, and the consequent stress pro- 
duced. 

§ 149. Greatest Stresses in Semi-Girder. — Wherever the 
section be assumed in a semi-girder, it is evident that any load 
placed on the truss at any point between the section and the 
free end increases both the shearing-force and the bending- 
moment at that section, and that any load placed between the 
section and the fixed end has no effect whatever on either 
the shearing-force or the bending-moment at that section. 

Hence every member of a semi-girder will have a greater 
stress upon it when the entire load is on, than with any partial 
load. 

§ 150. Greatest Chord Stresses in Girder supported at 
Both Ends. — Every load which is placed upon the truss, no 
matter where it is placed, will produce at any section whatever a 
bending-moment tending to turn the two parts of the truss on 
the two sides of the section upwards from the supports ; i.e., so 
as to render the truss concave upwards. 

Hence every load that is placed upon the truss causes com- 
pression in every horizontal upper chord, and tension in every 
horizontal lower chord. Hence, in order to obtain the greatest 
chord stresses, we assume the whole of the moving load to be 
upon the bridge. 

§ 151. Greatest Diagonal Stresses in Girder supported 
at Both Ends. — To determine the distribution of the load 
that will produce the greatest stress of a certain kind (tension 
or compression) in any given diagonal, let us suppose the diag- 
onal in question to be 7-8 (Fig. 109), through which we take 
our section ab. Now, it is evident that any load placed on the 
truss between ab and the left-hand (nearer) support will cause a 
shearing-force at that section which will tend to slide the part 
of the girder to the left of the section downwards with refer- 
ence to the other part, and hence will cause a compressive 
stress in 7-8 ; while any load between the section and the right- 



I 84 APPLIED MECHANICS. 

hand (farther) support will cause a shearing-force of the oppo- 
site kind, and hence a tension in the bar 7-8. 

Now, the bridge weight itself brings an equal load upon each 
joint ; hence, when the bridge weight is the only load upon the 
truss, the bar 7-8 is in tension. 

Hence, any load placed upon the truss between the section 
and the farther support tends to increase the shearing-force at 
that section due to the dead load (provided this is equally dis- 
tributed among the joints) ; whereas any load placed between 
the section and the nearer support tends to decrease the shear- 
ing-force at the section due to the dead load, or to produce a 
shearing-force of the opposite kind to that produced by the dead 
load at that section. 

Hence, if we assume the dead load to be equally distributed 
among the joints, we shall have the two following propositions 
true : — 

(a) In order to determine the greatest stress in any diagonal 
which is of the same kind as that produced by the dead load, 
we must assume the moving load to cover all the panel points 
between the section and the farther abutment, and no other 
panel points. 

(b) In order to determine the greatest stress in any diagonal 
of the opposite kind to that produced by the dead load, we must 
assume the moving load to cover all the panel points between 
the section and the nearer abutment, and no others. 

This will be made clear by an example. 

Example I. — Given the truss shown in Fig. 113. Length 

of chord = length of' diagonal = 

\AAAAAAA/k IO feet ' Dead load = 8oo ° lbs ' 

applied at each upper panel point. 
Moving load = 30000 lbs. applied 
at each upper panel point. Find 



10 12 u ie 

Fig. 



the greatest stresses in the members. 



EXAMPLE OF BRIDGE-TRUSS. 



185 



Solution, (a) Chord Stresses. — Assume the whole load to 
be upon the bridge : 



this will give 38000 
lbs. at each upper 
panel point ; i.e., omit- 
ting 1 and 17, where 
the load acts directly 
on the support, and 
not on the truss. 



+ 208423 (5) + 296181 (?) + 340059 (9) 




Fig. 114. 



Hence, considering the bridge so loaded, we shall have the fol- 
lowing results for the chord stresses : — 

Each supporting force = 38ooof-) = 133000. 



Section at 


Bending-Moment, in foot-lbs. 


2 


16 


133000 X 


5 


= 665000 


3 


15 


133000 X 


10 


= 1330000 


4 


14 


133000 x 


15 


— 38000 x 5 = 1805000 


5 


13 


133000 X 


20 


— 38000 X 10 = 2280000 


6 


12 


133000 X 


2 5 


- 3 8ooo( 5+15) = 2565000 


7 


11 


133000 X 


30 


— 38000(10 + 20) = 2850000 


8 


10 


133000 X 


35 


- 3 8ooo ( 5 + l 5 + 2 5) = 2945000 


9 




133000 X 


40 


— 38000(10 4- 20 4- 30) = 3040000 



Numbers of Chords. 


Stresses in Upper Chords. 


i-3 
3-5 

5-7 
7-9 

1 


15-17 

13-15 

11-13 

9-1 1 


665000 x 0.11547 = 4- 76788 
1805000 X 0.1 1547 = 4-208423 

2565000 x 0.11547 = 4-296181 
2945000 x 0.11547 = 4-340059 



1 86 



APPLIED MECHANICS. 



Numbers of Chords. 


Stresses in Lower Chords. 


2- 4 
4- 6 
6- 8 
8-io 


14-16 
12-14 
10-12 


-1330000 X O.11547 = -153575 

— 2280000 X 0.11547 = —263272 

— 2850000 x 0.1 1547 = —329090 

— 3040000 X 0.11547 = —351029 



Next, as to the diagonals, take, for instance, the diagonal 
7-8. When the dead load alone is on the bridge, the diagonal 
7-8 is in tension. From the preceding, we see that the greatest 
tension is produced in this bar when the moving load is on the 
points 9, 11, 13, and 15, and the dead load only on the points 3, 
5, 7. Now, a load of 38000 lbs. at 13, for instance, causes a 

shearing-force of —(38000) = 9500 lbs. at any section to the 

left of 13; and this shearing-force tends to cause the part to 

the left of the section to slide upwards, and that to the right 

downwards. 

On the other hand, with the same load at the same place, 

1 2 
there is produced a shearing-force of — (38000) = 28500 lbs. 

at any section to the right of 13 ; and this shearing-force tends 
to cause the part to the left to slide downwards, and that to the 
right upwards. Paying attention to this fact, we shall have, 
when the loads are distributed as above described, a shearing- 
force at the bar y-S causing tension in this bar ; the magnitude 
of this shearing-force being 



38000 
16 



(2+4 + 6 + 8) 



8000 
16 



(2 + 4 + 6) = 41500. 



Hence, we may arrange the work as follows : — 



GREATEST DIAGONAL STRESSES IN GIRDER. 



1 8/ 









Greatest 
Stresses in 


Numbers of 
Diagonals. 


Greatest Shearing-Forces producing Stresses of Same Kind as 
Dead Load. 


Diagonals of 

Same Kind 

as those due 

to Dead 

Load. 


1—2 


17-16 


^(2 + 44-6+8+10+12+14) 


= 133000 


-153575 


2-3 


16-15 


^(2+4+6+8+ 10+ 12+ 14) 


= 133000 


+ 153575 


3-4 


15-M 


3^ (2 + 4+6+8+IO+I2) _^ (2) 


= 98750 


— 1 14027 


4-5 


14-13 


^ + 4+6+8+10+12) -^(2) 


= 98750 


+ I 14027 


5-6 


13-12 


^(2+4+6+8+10) -^(2+4) 


= 68250 


— 78808 


6-7 


1 2-1 1 


^(2+4+6+8 + 10) -^°(2+ 4 ) 


= 68250 


+ 78808 


7-8 


II-IO 


^(2+4+6+8) - ^(2 + 4+6) 


— 41500 


— 47920 


8-9 


10- 9 


^(2+4+6+8) -^°(2+ 4 +6) 


— 4^00 


+ 479 2 







Greatest 






Stresses in 


Numbers of 


Greatest Shearing-Forces producing Stresses of Kind Opposite 


Diagonals of 


Diagonals. 


from Dead Load. 


Kind Oppo- 
site from 
Dead Load. 


8-9 


10- 9 


^?( 2 +4+6) - ^( 2+4 +6+8) = .8500 


—21362 


7-S 


II-IO 


«fVU) -^+4+6+8+.o) =- 75 o 


None. 



Hence, the diagonals 8-9 and 9-10 are the only ones that, 
under any circumstances, can have a stress of the kind oppo- 
site to that to which they are subjected under the dead load 
alone. 



1 88 APPLIED MECHANICS. 

Fig. 1 14 exhibits the manner of writing the stresses on the 
diagram. 

§152. General Application of this Method. — It is plain 
that the method used above will apply to any single system of 
bridge-truss with horizontal chords and diagonal bracing, what- 
ever be the inclination of the braces. 

When seeking the stress in a diagonal, the section must be 
so taken as to cut that diagonal ; and, as far as this stress alone 
is concerned, it may be equally well taken at any point, as well 
as near a joint, provided only it cuts that diagonal which is in 
action under the load that produces the greatest stress in this 
one, and no other. 

On the other hand, when we seek the stress in a horizontal 
chord, the section might very properly be taken through the 
joint opposite that chord. 

Taking it very near the joint, only serves to make one sec- 
tion answer both purposes simultaneously. 

§ 153. Bridge-Trusses with Vertical and Diagonal Bra- 
cing. — When, as in Figs, in and 112, there are both vertical 
and diagonal braces, and also horizontal chords, we may deter- 
mine the stresses in the diagonals and in the chords just as 
before ; only we must take the section just to one side of a joint, 
and never through the joint. 

As to the verticals, in order to determine the stress in any 
vertical, we must impose the conditions of equilibrium between 
the vertical components of the forces acting at one end of that 
vertical: thus, if the loads are at the upper j.oints in Fig. ill, 
then the stress in vertical 3-2 must be equal and opposite to 
the vertical component of the stress in diagonal 1-2, as these 
stresses are the only vertical forces acting at joint 2. 

Vertical 5-4 has for its stress the vertical component of the 
stress in 3-4, etc. Thus 

Stress in 3-2 = shearing-force in panel 1-3, 
Stress in 5-4 = shearing-force in panel 3-5, etc. 



TRUSSES WITH VERTICAL AND DIAGONAL BRACING. 1 89 

On the other hand, if the loads be applied at the lower 
joints, then 

Stress in 3-2 = shearing-force in panel 3-5, 
Stress in 5-4 = shearing-force in panel 5-7, etc. 

Example. — Given the truss shown in Fig. in. Given 
panel length = height of truss = 10 feet, dead load per panel 
point == 12000 lbs., moving load per panel point = 23000 lbs. ; 
load applied at upper joints. 

Solution, (a) Chord Stresses. — Assume the entire load on 
the bridge, i.e., 35000 lbs. per panel point. Hence 

Total load on truss =13 (35000) = 455000 lbs., 
Each supporting force = 227500 lbs. 



Joint near 






which 
Section is 


Bending-Moment at the Section very near the Joint, on Either Side of the Joint. 


taken. 






I 


28 







3 


27 


227500 X 10 


= 2275000 


5 


25 


227500 X 20 — 35000 X 10 


= 4200000 


7 


23 


227500 x 30 — 35000(10 4- 20) 


= 5775000 


9 


21 


227500 x 40 — 35000 (10 + 20 + 30) 


= 7OO000O 


11 


19 


227500 X 50 — 350OO (IO + 20 + 30 + 40) 


= 787500O 


13 


17 


227500 x 60 — 35000(10 + 20 -f- 30 + 40 + 50) 


= 840OOOO 


15 


— 


• 227500 x 70 — 35000(10 + 20 + 30 -f 40 -t- 50 + 60) 


= 85750OO 



To find any chord stress, divide the bending-moment at a 
section cutting the chord, and passing close to the opposite 
joint, by the height of the girder, which in this case is 10. 
Hence we have for the chord stresses (denoting, as before, com- 
pression by -f, and tension by — ) : — 



i go 



APPLIED MECHANICS. 



Stresses in Upper Chords. 


Stresses in Lower Chords. 


i- 3 

3" 5 
5" 7 
7" 9 
9-1 1 
11-13 

i3- J 5 


27-28 

25-27 

23-25 
21-23 

19-21 

17-19 

i5-!7 


+ 227500 
+420000 

+ 5775°° 
+ 700000 

+ 787500 

+ 840000 

+857500 


2- 4 
4- 6 
6- 8 
8-10 
10-12 
12-14 


24-26 
22-24 
20-22 
18-20 
16-18 
14-16 


— 227500 

— 420000 

-577500 

— 700000 
-787500 

— 840000 



Diagonals. — It is evident, that, for the diagonals, the same 
rule holds as in the case of the Warren girder : i.e., the greatest 
stress of the same kind as that produced by the dead load 
occurs when the moving load is on all the joints between the 
diagonal in question and the farther abutment ; whereas the 
greatest stress of the opposite kind occurs when the moving 
load covers all the joints between the diagonal in question and 
the nearer abutment. 

The work of determining the greatest shearing-forces may 
be arranged as in tables on p. 191. 

Counterbraces. — If the truss were constructed with those 
diagonals only that slope downwards towards the centre, and 
which may be called the main braces, the diagonals n-12, 
13-14, 14-17, and 16-19 would sometimes be called upon to 
bear a thrust, and the verticals 12-13 and 17-16 a pull: this 
would necessitate making these diagonals sufficiently strong 
to resist the greatest thrust to which they are liable, and fixing 
the verticals in such a way as to enable them to bear a pull. 

In order to avoid this, the diagonals 10-13, 12-15, 15-16, 
and 17-18 are inserted, which are called counterbraces, and 
which come into action only when the corresponding main 



TRUSSES WITH VERTICAL AND DIAGONAL BRACING. 191 

braces would otherwise be subjected to thrust. They also 
prevent any tension in the verticals. 



Diagonals. 


Greatest Shearing 


-Forces of the Same Kind as those produced by- 
Dead Load. 


I- 2 


28-26 


«a*,+,+3+. 


..+13) =227500 


3- 4 


27-24 


3 -f°(. + z + 3+. 


12000/ N 

- - + l2 ) - -^-i 1 ) = T 94i43 


5-6 


25-22 


^(. + 2 + 3 +. 


. I2COO, . . s 

..+11) -— (l + 2) = 162429 


7-8 


23-20 


^fC + 2 + 3+. 


. 12000, , , . 

• • + iq )- "^r( I+2+ 3) = 132357 


9-10 


21-18 


^(■ + 2 + 3+. 


• •+ 9) - x -^p(i+ 2+...+ 4)= 103929 


11-12 


19-16 


^(■ + 2+3+- 


••+ 8)-^°(i + 2-|-...+5)= 77143 


13-14 


17-14 


S2?(, + 2 + 3 +. 


••+ 7)- ~(r+2+.. .+6)= 52000 



Diagonals. 



Greatest Shearing-Forces of the Opposite Kind to those produced by- 
Dead Load. 



13-14 

11-12 
9-IO 



17-14 
I9-I6 
2I-I8 



^(r + 2 + 3+... + 6)-^-°(i + 2+... + 7) 
14 

.. + 5)-^pe+2+...+8) 

■• + 4) -==*M-2+.... + 9 ) 



^(. + 2+ ... + 5) -==fr+a+'...+ 

14 14 



(r + 2 + 



28500 

6643 

-I3S7I 



The main braces and counterbraces of a panel are never in 
action simultaneously. Hence we have, for the greatest stresses 
in the diagonals, the following results, obtained by multiplying 

the corresponding shearing-forces by 



1 
cos 45' 



= 1.414. 



192 



APPLIED MECHANICS. 



In the following I have used this number to three decimal 
places, as being sufficiently accurate for practical purposes. 



Stresses in Main Braces. 


Stresses in Counterbraces. 


I- 2 


28-26 


-321685 


15-12 


I5-I6 


-40299 


3" 4 


27-24 


-274518 


I3-IO 


I7-I8 


- 9393 


5" 6 


25-22 


— 229675 








7" 8 


23-20 


-187153 








9-10 


21-18 


-146956 








11-12 


19-16 


— 109080 








13-H 


17-14 


- 73528 









Vertical Posts. — Since the loads are applied at the upper 
joints, the conditions of equilibrium at the lower joints require 
that the thrust in any vertical post shall be equal to the vertical 
component of the tension in that diagonal which, being in action 
at the time, meets it at its lower end. 

Hence it is equal to the shearing-force in that panel where 
the acting diagonal meets it at its lower end. 

The post 15-14 is a counterbrace, and comes into action 
only when the diagonal counterbraces are in action. 

We therefore have, for the posts, the following as the greatest 
thrusts : — 

STRESSES IN VERTICALS. 



3" 2 


27-26 


+ 227500 


5" 4 


25-24 


+•194143 


7- 6 


23-22 


-I-162429 


9- 8 


21-20 


+ 1 3 2 357 


II-IO 


19-18 


+ 103929 


13-12 


17-16 


+ 77 T 43 


r 5" 


-14 


+ 28500 



CONCENTRATING THE LOAD AT THE JOINTS. 



193 



t 





Fig. 115 shows the stresses marked on the diagram. 

§ 154. Manner of Concentrating the Load at the Joints 

— In using the methods given above, we are 

assuming that all the loads are concentrated 

at the joints, and that none are distributed 

over any of the pieces. As far as the mov- 
ing load is concerned, and also all of the 

dead load except the weight of the truss 

itself, this always is, or ought to be, effected ; 

and it is accomplished in a manner similar 

to that adopted in the case of roof-trusses. 

This method is shown in the figure (Fig. 

116); floor-beams being laid across from 
girder to girder at the joints, 
on top of which are laid longi- 
tudinal beams, and on these 
the sleepers if it is a railroad 
bridge, or the floor if it is a 
road bridge. The weight of 
the truss itself is so small a 
part of what the bridge is 
called upon to bear, that it 
can, without appreciable error, 
be considered as concentrated 
at the joints either of the up- 
per chord, of the lower chord, 
or of both, according to the 
manner in which the rest of 
the load is distributed. 

§ 155. Closer Approxima- 
tion to Actual Shearing- 
Force. — In our computations 
of greatest shearing-force, we 






Fig. 116. 



Fig. 115. 



make an approximation which is generally considered to be 



194 APPLIED MECHANICS. 

sufficiently close, and which is always on the safe side. To 
illustrate it, take the case of panel 3-5 of the last example. 
In determining its greatest shearing-force, we considered a load 
of 35000 lbs. per panel point to rest on all the joints from the 
right-hand support to joint 5, inclusive, and the dead load to 
rest on all the other joints of the truss. Now, it is impossible, 
if the load is distributed uniformly on the floor of the bridge, 
to have a load of 35000 lbs. on 5 and 12000 on 3 simultaneously ; 
for, if the moving load extended on the bridge floor only up to 
5, the load on 5 would be only 12000 + £(23000) = 23500 lbs., 
and that on 3 would then be 12000 lbs. If, on the other hand, 
the moving load extends beyond 5 at all, as it must if the load 
on 5 is to be greater than 23500 lbs., then part of it will rest 
on 3, and the load on 3 will then be greater than 12000 lbs. ; 
for whatever load there is between 3 and 5 is supported at 
3 and 5. 

Moreover, we know that the effect of increasing the load on 
5 is to increase the shearing-force, provided we do not at the 
same time increase that on 3 so much as to destroy the effect 
of increasing that on 5. 

Hence, there must be some point between 3 and 5 to which 
the moving load must extend in order to render the shearing- 
force in panel 3-5 a maximum. 

Let the distance of this point from 5 be x ; then, if we let 

2 3°°° • r r 1 

w = — — — = moving load per toot of length, 

Moving load on panel = wx, 

uuX^ 

Part supported at 3 = , 

20 

idX* 

Part supported at 5 = wx — 



20 



Hence, portion of shearing-force due to the moving load on 

panel 3-5 equals 



CONCENTRATING THE LOAD AT THE JOINTS. 



195 



12/ WX 2 \ I wx 2 w I \xx*\ 

( WX — ) = — { I 2X ). 

i4\ 20 / 14 20 14V 20 / 

This becomes a maximum when its first differential co-efficient 
becomes zero, i.e., when 

12 - \%x = o; 



therefore 



x = 9 .23. 



Hence, when the moving load extends to a distance of 9.23 feet 
from 5, then the shearing-force in panel 3-5, and hence the 
stress in diagonal 3-4, is a maximum. 



Diagonals in Panel 
where Shearing- 
Force is taken. 


Portion of Shearing-Foxce 

due to Moving Load on 

Panel. 


Value 

of x, in 

feet. 


Portion of Load 

at Joints named 

below. 


Portion of Load 

at Joints named 

below. 


1- 3 


27-28 


wf 13^ 

— (l3* - -^— J 

I4\ 20 / 


IO.OO 


I 


1 1 500 


3 


1 1 500 


3- 5 


25-27 


w / 13.2A 
14V 20 / 


9- 2 3 


3 


9797 


5 


1 1432 


5- 7 


23-25 


I4\ 20 / 


8.46 


5 


8230 


7 


1 1 227 


7- 9 


21-23 


e( 10x - ^) 

14V 20 / 


7.69 


7 


6801 


9 


10886 


9-1 1 


19-21 


14V 20 / 


6.92 


9 


5507 


11 


10409 


11-13 


17-19 


.14 V 20 / 


6.15 


11 


435o 


13 


9795 


13-15 


15-17 


Z( 1x - I3£?) 

I4\ 20 / 


5.38 


13 


3329 


15 


9°45 ( 



To show how the adoption of this method would affect the 
resulting stresses in the diagonals and verticals, I have given 
the work above, and shown the difference between these and 



196 



APPLIED MECHANICS. 



the former results. In this table x = distance covered by load 
from end of panel nearest the centre. 



Panels. 


Greatest Shearing-Force of Same Kind as that due to Dead Load. 


1- 3 


27-28 


35000 . 

— (1 + ...+ 13) =227500 


3- 5 


25-27 


^(i+...+ ii)+^(23432)~(2i 7 97) =183528 


5" 7 


23-25 


35000, . . ..11. , 2 , ,i, 

— (I+. . .+ IO) + -(23227)— -(2023O) --(1 2000) =152003 


7- 9 


21-23 


3 -^(i+...+ 9)+^(22886)-f 4 (i88oi)-^°(i + 2) =122247 


9-1 1 


19-21 


^P(i+...+ 8)+^(22409}-^(i7507)-^°(i + 2+3) = 94261 


11-13 


17-19 


^f(i+...+ 7 )+7 4 (2i795)-r 4 ( l6 35o)-^°(i + ..-+4)= 68043 


13-15 


15-17 


3 -fP(i+...+ 6)+f 4 (2i66 4 )-^(i 4 7io)-l-^P(i+...+5)= 44171 



Hence, for stresses in main braces, we have 



Diagonals. 


Stresses. 


I- 2 


28-26 


-321685 


3- 4 


27-24 


- 2 595°9 


5-6 


25-22 


-214932 


7-8 


23-20 


-172857 


9—10 


21-18 


-133285 


11-12 


19-16 


- 96213 


13-M 


17-14 


- 62458 



Moreover, for the shearing-forces of opposite kind from 



CONCENTRATING THE LOAD AT THE JOINTS. 



197 



those due to dead load, we have, if ^r = distance from end of 
panel nearest support which is covered by moving load, — 



Panels. 


Portion of Shear due 
to Moving Load on Panel. 


Value 
of*. 


Portion of Load 

at Joints named 

below. 


Portion of Load 

at Joints named 

below. 


13-15 
II-I3 


17-15 
19-17 


™( 6x - s?} 
14V 20 / 

14V 20 / 


4.62 
3.84 


IS 

13 


2455 
169s 


13 
II 


8I7I 
7137 



Panels. 



Greatest Shearing-Forces of Opposite Kind from those due to Dead Load. 



13-15 
II-13 



17-15 
19-17 



3 -^(i + ...+5)+r 4 (20i7i)-f 4 (i4455)--~(i+..-+6) = 20917 
^(i+. . .+4)+f 4 (i9i37)-^(i3695)-^°(i+. ■ .+7) = 1965 



Hence we have the following as the stresses in the counter- 
braces : — 



Counter-Braces. 


■ Stresses. 


15-12 
I3-IO 


T5-16 
17-18 


-29577 
- 2778 



And, for the verticals, we have the new, instead of the old, 



shearing-forces. 



93 



APPLIED MECHANICS. 



The following table compares the results : — 



Diagonals. 


Stress, Ordinary 
Method. 


Stress, New Method. 


Difference. 


I- 2 


28-26 


-321685 


-321685 




3" 4 


27-24 


-274518 


-2595°9 


I5OO9 


5" 6 


25-22 


-229675 


-214932 


14743 


7" 8 


23-20 


-187153 


-I72857 


I4296 


9-10 


21-18 


— I46596 


- J 33285 


I367I 


11-12 


19-16 


— IO9080 


- 96213 


I2867 


i"3-' I 4 


17-14 


- 73528 


- 62458 


IIO7O 


15-12 


15-16 


- 4O299 


- 29577 


IO722 


13-10 


17-18 


- 9393 


- 2783 


761O 



Verticals. 


Stress, Ordinary 
Method. 


Stress, New Method. 


Difference. 


3" 2 


27-26 


+ 2275OO 


+ 2275OO 


I0615 


5- 4 


25-24 


+ I94M3 


+ 183528 


IO426 


7- 6 


23-22 


-f-162429 


+ I52OO3 


IOIIO 


9-8 


21-20 


+ 132357 


+ 122247 


9668 


II-IO 


19-18 


+ IO3929 


+ 94261 


9IOO 


13-12 


17-16 


+ 77143 


+ 68043 


7583 


J 5- 


-14 


+ 285OO 


+ 20917 





§156. Compound Bridge-Trusses. — The trusses already 
discussed have contained but a single system of latticing, or 



COMPO UND BR ID GE- TR USSES. 



I 99 



at least only one system that comes in play at one time ; so that 
a vertical section never cuts more than three bars that are in 
action simultaneously, the main brace having no stress upon it 
when the counterbrace is in action, and vice versa. 

We may, however, have bridge-trusses with more than one 
system of lattices ; and, in determining the stresses in their 
members, we must resolve them into their component systems, 
and determine the greatest stress in each system separately, 
and then, for bars which are common to the two systems, add 
together the stresses brought about by each. 

In some cases, the design is such that it is possible to 
resolve the truss into systems in more than one way, and then 
there arises an uncertainty as to which course the stresses will 
actually pursue. 

In such cases, the only safe way is to determine the greatest 
stress in each piece with every possible mode of resolution of 
the systems, and then to design each piece in such a way as to 
be able to resist that stress. 

Generally, however, such ambiguity is an indication of a 
waste of material ; as it is most economical to put in the bridge 
only those pieces that are absolutely necessary to bear the 
stresses, as other pieces only add so much weight to the struc- 
ture, and are useless to bear the load. 

The mode of proceeding can be best explained by some 
examples. 



Example I. 



Given the lattice-girder shown in Fig. 



I7> 



5 7 9 11 13 15 17 19 21 23 25 



^xxxxxxxm^ 



4 6 8 10 12 14 16 18 20 22 24 20 



loaded at the lower panel points 

only. Dead load = 7200 lbs. 

per panel point, moving load 

= 18000 lbs per panel point; 

let the entire length of bridge 

be 60 feet ; let the angle made by braces with horizontal 

= 6o°. 



Fig. 



200 



APPLIED MECHANICS. 



This truss evidently consists of the two single trusses shown 
in Figs. 1 1 ya 



1 


5 9 13 17 21 25 




\/\A/\AA/ 




s 


4 


8 12 16 20 24 


N 



19 23 



22 



Fig. 



and wyb ; and 
we can compute 
the greatest 
stress of each FlG - I1?a ' 

kind in each member of these trusses, and thus 
obtain at once 
all the diag- 
onal stresses, 
and then, by 
addition, the 
greatest chord stresses. 

Thus the stress in 1-3 (Fig. 117) is the 
same as the stress in 1-5 (Fig. 11 yd). 

The stress in 3-5 = stress in 1-5 (Fig. 
nya) -J- stress in 3-7 (Fig. 117^). 

The stress in 5-7 = stress in 5-9 (Fig. 
11 y a) -f- stress in 3-7 (Fig. 117^). 

The results are given on the diagram (Fig. 
117c) ; the work being left for the student, as 
it is similar to that done heretofore. 



II. — Given the 



lattice-girder 



Fig. 117c 



Example 

shown in Fig. 118. Given, as before, Dead 
load = 7200 lbs. per panel point, moving load 
= 18000 lbs. per panel point, entire length of 
bridge — 25 feet ; load applied at lower panel 
points. 



Solution. — In this case, there are two possible modes of 
resolving it into systems. The first is shown in Figs. 118a and 
\\%b: and this is necessarily the mode of division that must 
hold whenever the load is unevenly distributed, or when the 



COMPOUND BRIDGE-TRUSSES. 



201 



travelling-load covers only a part of the bridge ; for a single 
load at 6 is necessarily put in communication with the support 
at 2 by means of the diagonals 6-3 and 3-2, and with the sup- 
port at 3 by means of the diagonals 6-7, 7-10, 10-11, and the 
vertical 11-12, and can cause no stress in the other diagonals. 



1 3 5 7 9 11 



2 4 6 8 10 12 
Fig. 118. 



2 6 10 12 

Fig. ii8«. 



2 4 8 12 

Fig. ii8<5. 



11 



I 57 

24 10 12 

Fig. ii8<t. 



2 6 8 12 

Fig. ii&/. 



When, however, the whole travelling-load is on the bridge, 
it is perfectly possible to divide it into the two trusses shown 
in Figs. 11&: and n8d, the diagonals 4-5, 7-10, 6-7, and 5-8 
having no stress upon them. 

When the load is unevenly distributed, we have certainly 
the first method of division ; and when evenly, we are not sure 
which will hold. 

Hence we must compute the greatest stresses with each 
mode of division, and use for each member the greatest ; for 
thus only shall we be sure that the truss is made strong 
enough. 

We shall thus have-the following results : — 



202 



APPLIED MECHANICS. 



FIRST MODE OF DIVISION (FIGS. \\Za AND \\U). 



Diagonals. 


Greatest Shearing-Force 
of One Kind. 


Greatest Shearing-Force 
of Opposite Kind. 






Fig. 
118a. 


Fig. 
1 1 83. 




2 - 3 
3-6 
6-7 
7-10 

IO-II 


12-9 
9-8 
8-5 
5-4 
4-1 


25200. . . ,, 

-7- (3+ = 20160 

^(3+ 1) = 20160 

25200 

— = 5040 

25200 

— = 5040 

=0 



O 

53=5(8) = 10080 

^°(2) = JOOSO 

25200 . 

-^-(2 + 4) =3 240 


+ 23279 
-23279 
+ 582O 
— 582O 
O 


— O 
+ O 
-1 1639 
+ I 1639 

-349 1 8 



Chords. 
Supporting force at 2 (Fig. uSa) or 12 (Fig. n85) 

= ^°( 3 + 1) = 20160, 
Supporting force at 12 (Fig. 1180) or 2 (Fig. u8£) 

= ^(2+4)^30240. 



Section. 




Chords. 


Maxi- 
mum 




Components 


Greatest 










Chords. 


of 


Resultant 








<3 
00 

bi 


00 

bi 

E 


in 
Separate 
Trusses. 




Stresses. 


Stresses. 


3 


9 


20160 X 5=iooSoo 


2- 6 


8-12 


-1 1639 


i-3 


9-1 1 


0+I-5 


+ 17459 


6 


8 


20160X 10=201600 


3" 7 


5- 9 


+ 23279 


3-5 


7- 9 


3" 7 + 1-5 


+ 40738 


7 


5 


30240 X 10=302400 


6-10 


4-8 


~349 lS 


5-7 




3- 7 + 5-9 


+ 46558 


10 


4 


30240 x 5=151200 


7-1 1 


'- 5 


+ 17459 


2-4 


10-12 


2- 6+2-4 


— 1 1639 








10-12 


2- 4 





4-6 
6-S 


8-10 


2- 6+4-S 
6-10+4-S 


-46557 
-69S36 



COMPOUND BRIDGE-TRUSSES. 



203 



SECOND METHOD OF DIVISION (FIGS. 11SV AND \\%d). 

Diagonals (Fig. 118^). 



Diagonals. 


Maximum 
Shear. 


Corresponding 
Stresses. 


1-4 

4-5 


IO-II 

7-10 


252OO 
O 


— 29O98 
O 



Fig. \\%d. 



Diagonals. 


Maximum 
Shear. 


Corresponding 
Stresses. 


2-3 


9-12 


252OO 


+ 29O98 


3-6 


8-9 


252OO 


— 29O98 


6-7 


5-8 


O 


O 



Chords. 

Each supporting force in either figure = 25200. 
Fig. 1 18c. 

Bending-moment anywhere between 4 and 10 = (25200) (5) = 126000; 

.*. Stress in 1-11 = +14549, 
.*. Stress in 4-10 = —14549. 

Fig. nSd. 

Bending-moment at 3 or 9 = 126000, 

Bending-moment anywhere between 6 and 8 = 252000; 

.*. Stress in 3-9 = +29098, 

Stress in 2-6 or 8-12 = —14549, 
Stress in 6-8 = —29098. 



204 



APPLIED MECHANICS. 



Hence we have for chord stresses, with this second divis- 



ion, 



Chords. 




Stresses. 


i-3 


9-1 1 


I-II 4- o 


+ 14549 


• 3-5 


7- 9 


i-ii + 3-9 


+ 43647 


5-7 


- 


i-ii + 3-9 


+ 43647 


2-4 


IO-I2 


o + 2-6 


-14549 


4-6 


8-io 


4-10 + 2-6 


— 29O98 


6-8 


- 


4-10 4- 6-8 


-43647 



Hence, selecting for each bar the greatest, we shall have, as 
the stresses which the truss must be able to resist, — 



1-4 


10-11 


+ 


-34918 


i-3 


9-1 1 


+ 17459 


2 -3 


12-9 


+ 29098 


— 


3-5 


7" 9 


+43647 


3-6 


9-8 


+ 


— 29098 


5-7 


- 


+46558 


4-5 


10- 7 


+ 11639 


- 5 82 ° 


2-4 


10-12 


-14549 


5-8 


7" 6 


+ 5820 


-1 1639 


4-6 
6-8 


8-10 


-46557 
— 69836 



These results are recorded in Fie:. 1 18^. 

(D+17549 (3)+ lJK47(5)+4G55SCtt+ 43647(9 )+lT549(ifl 




' (2^-44549(4)— 4C558 (C)-G'J83G(8)-46558(10)-14549(12) 
Fig. 118*. 

§157. Other Trusses. — In Figs. 119, 120, and 121, we 
have examples of the double-panel system with the load placed 



OTHER TRUSSES. 



205 



at the lower panel points only. When, as in 119 and 120, the 
number of panels is odd, the same ambiguity arises as took place 
in Fig. 118. When, on the other hand, the number of panels 
is even, as shown in Fig. 121, there is only one mode of division 
into systems possible. The diagrams speak for themselves, and 
need no explanation. 



2 4 6 8 10 '12 14 16 18 20 22 24 26 28 30 32 




2 4 



12 16 20 24 28 32 





17 21 25 
Fig. uga. 



26 30 32 




31 33 34 



2 4 




22 26 30 32 




17 19 23 27 31 33 31 
Fig. 119c. 



24 28 





11 15 21 25 29 33 34 

Fig. i 19a?. 



2C6 



APPLIED MECHANICS. 



2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 




2 4 



12 16 20 24 28 32 36 



\ 






\ 


X / 

\ / 

/ \ 

y X 


\ / 
/ \ 






/ 



1 3 



15 19 23 

Fig. 120a. 



27 31 



2 6 10 14 18 



26 SO 34 








1 5 9 13 



17 21 

Fig. i2o3. 



25 29 33 3d 



2 6 10 14 



24 28 S2 



1 5 9 13 



17 19 
Fig. 120c. 



27 31 35 



2 4 



26 30 34 36 



21 25 29 33 35 



Fig. laod. 



2 4 6 8 10 12 14 16 18 






1 8 5 7 9 11 13 15 17 19 80 
Fig. iaof. 



FINK'S TRUSS. 



207 




2 4 



12 16 18 




Fig. i2i<5. 



Fig. 122. 



The trusses given above may be considered as examples, to 
be solved by the student by assuming the dead and the moving 
load per panel point respectively. 

§158. Fink's Truss. — The description of this truss will 
be evident from the figure. There is, first, the primary truss 
1-8-16 ; then on each side 
of 9-8 (the middle post of 
this truss) is a secondary 
truss (1-4-9 on tne l e ft> 
and 9-1 2-16 on the right). 

Each of these secondary 
trusses contains a pair of smaller secondary trusses, and the 
division might be continued if the segments into which the 
upper chord is thus divided were too long. 

Of the inclined ties, there is none in which any load tends 
to produce compression ; in other words, every load either in- 
creases the tension in the tie, or else does not affect it. Hence 




Fig. 123. 



208 



APPLIED MECHANICS. 



the greatest stresses in all the members will be attained when 
the entire travelling-load is on the truss, and we need only con- 
sider that case. 

The determination of the stress in any one member can 
readily be obtained by determining, by means of the triangle 
of forces, the stress in that member due to the presence of 
the total load per panel point, at each point, and then adding the 
results. This will be illustrated by a few diagonals. 

Let angle 8-1-9 — h 
Let angle 4-1-5 = i„ 
Let angle 2-1-3 — ** '> 

we shall have, if w + w , = entire load per panel point, — ■ 



c 
. 

•£ in 

a ° 


Effect of Loads at 


C 
*> O 


3 


5 


7 


9 


11 


13 


15 


1-2 

2-5 
5-6 

6- 9 
i-4 
4-9 
i-S 


zu 4- Wi 
2 sin i 2 

2 sin i 2 





ZV 4" ZU t 

4 sin z'i 

zo 4- iv 1 
4 sin ^ 

ZV 4" Wj 

8 sin / 








zu 4- zu 1 
2 sin z'i 

ZU + 7X/t 

2 sin *i 

tx; 4- w r 
4 sin t 





W 4- 7f r 

2 sin 4 

Ttf 4~ W t 

2 sin z 2 

ZU 4" Z0i 

4 sin fj 

w 4- Wi 
4 sin *! 

Z0 + 7X/i 

S sin » 











zu + W\ 
2 sin / 









ZV 4" ZUi 

8 sin i 













zu 4- zut 
4 sin i 


O 







zu 4- u'i 
8 sin i 


7tV 4" K'i 


2 sin / 2 
7c; 4- rv t 


2 sin » a 

70 4" ZU x 

2 sin r 2 

W 4" 7£/ t 

2 sin i 9 

ZV 4" 7^ 

sin r'i 

zu 4- «'i 

sin r*! 

3(w+w x ) 

2 sin / 



The stresses in all the other members may be found in a 
similar manner. 



GENERAL REMARKS. 



20q 



§ 159. Bollman's Truss. — The description of this truss is 
made sufficiently clear by the figure. The upper chord is made 
in separate pieces ; and 
the short diagonals 2-5, 
3-4, 4-7, 5-6, 7-8, 6-9, 
8-1 1, and 9-10 are only 
needed to prevent a 
bending of the upper 
chord at the joints. 
When this is their only object, the stress upon them cannot be 
calculated : indeed, it is zero until bending takes place ; and 
then it is the less, the less the bending. Hence, in this case, 
the stress is wholly taken up by the principal ties ; and these 
have their greatest stress when the whole load is on the bridge. 

The computation of the stresses is made in a similar man- 
ner to that used in the Fink. 




Fig. 124. 



§160. General Remarks. — The methods already explained 
are intended to enable the student to solve any case of a bridge- 
truss where there is no ambiguity as to the course pursued by 
the stresses. 

In cases where a large number of trusses of one given type 
are to be computed, it would, as a rule, be a saving of labor to 
determine formulas for the stresses in the members, and then- 
substitute in these formulas. 

Such formulae may be deduced by using letters to denote 
the load and dimensions, instead of inserting directly their 
numerical values ; and then, having deduced the formulas for 
the type of truss, we can apply it to any case by merely sub- 
stituting for the letters their numerical values corresponding 
to that case. 

Such sets of formulas would apply merely to specific styles 
of trusses, and any variation in these styles would require the 
formulas to be changed. 



2IO APPLIED MECHANICS. 

In order to show how such formulas are deduced, a few will 
be deduced for such a bridge as is shown in Fig. ill, 

Let the load be applied at the upper panel points only ; let 
dead load per panel point = w, moving load per panel point 
= w x . Let the whole number of panels be N, N being an even 
number. Let the length of one panel = height of truss = /. 
Then length of entire span = NL 

Consider the n th panel from the middle. 

The stress in the main tie is greatest when the moving load 
is on all the panel points from the farther abutment up to the 
panel in question. 

Hence, for the n th panel from the middle, the greatest shear- 
ing-force that causes tension in the main tie 

w+wi ( N ) w ( N ) 

= ^H I+2+ 3+-"+7 + "[-M I+2 +3+-+7-*-- I | 

i ( vin V N 1 » ) 

= ^rlu+'v + 7 +«j+-( 2 «+i)^}. 

Hence stress in main tie 

= m{ w {{i + "J + t + "] + 1 (2n + 1)N }- (I) 

For the counterbrace, we should obtain, in a similar way, the 
formula 



si 2 j VfN V N 1 



wN , , ) 

~ (a«+ Of 



which represents tension when it is positive. Proceed in a 
similar way for the other members. 

When there is more than one system, we must divide the 
truss into its component systems ; and when there is ambiguity, 
we must use, in determining the dimensions of each member, 
the greatest stress that can possibly come upon it. 



CENTRE OF GRAVITY. 211 



CHAPTER V. 
CENTRE OF GRAVITY. 

§ 161. The centre of gravity of a body or system of bodies, is 
that point through which the resultant of the system of parallel 
forces that constitutes the weight of the body or system of 
bodies always passes, whatever be the position in which the 
body is placed with reference to the direction of the forces. 

§ 162. Centre of Gravity of a System of Bodies. — If 
we have a system of bodies whose weights are W x , W 2 , W v etc., 
the co-ordinates of their individual centres of gravity being 
(■^u yu <0> (^ 2 » j^2» #*) i (* v y v # 3 )» etc., respectively, and if we 
denote by x oi y , z Q , the co-ordinates of the centre of gravity of 
the system, we should obtain, just as in the determination of the 
centre of any system of parallel forces, — 

i°. By turning all the forces parallel to OZ, and taking 
moments about OY, 

(W l + W 2 4- W 3 4- etc.)* = W t x t 4- W 2 x 2 4- W 3 x 3 -f etc., 
or 

and, taking moments about OX, 

(W x 4- W 2 4 W 3 4- etc.)j>o = W x y x -f- 0^ a 4- W 3 y 3 + etc., 
or 



212 



APPLIED MECHANICS. 



2°. By turning all the forces parallel to OX, and taking 
moments about OY, 



or 



(W x 4- W 2 + W z + etc.K = W lZl + W 2 z 2 + ^ + etc., 



s Q 2^ = 5JF*. 



Hence we have, for the co-ordinates of the centre of gravity 
of the system, 

%Wx %Wy %Wz 

yo = 



%w 



%w 



^w 



EXAMPLES. 

I. Suppose a rectangular, homogeneous plate of brass (Fig. 125), 

where AD = 1 2 inches, AB = 5 inches, 
and whose weight is 2 lbs., to have 
weights attached at the points A, B, C, 
and D respectively, equal to 8, 6, 5, and 
x 3 lbs. ; find the centre of gravity of the 
system. 



D 




Y 


A 







f 












C 






B 



Fig. 125 

the centre of the rectangle, and we have 

W t == 2, W 2 = 8, IV, = 6, W A 

— £ * 



Solution. 
Assume the origin of co-ordinates at 



m 



5, " s = 3, 
x x =0, x 2 = 6, x 3 = 6, # A = — 6, x s = — 6, 

Jl\ =0, ;- 2 = f , j' 3 = -f, ;> 4 -f, r 5 = j; 

.-. 2/"#ff = o 4- 48 + 36 — 30.0 — 18.0 = 36, 
^Wy = o -f- 20 — 15 — 12.5 + 7.5 = o, 
%W =2-+- 8 4- 6+ 5.04- 3-0=24; 



3* 

24 



= i-5> )'o 



24 



= o. 



Hence the centre of gravity is situated at a point E on the line OX, 
where OE — 1.5. 




CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 21 1 

2. Given a uniform- circular plate of radius 8, and weight 3 lbs. 
(Fig. 126). At the points A, B, C, and D, 
weights are attached equal to 10, 15, 25, and 23 
lbs. respectively, also given AB = 45 , BC — 
105 , CD — 120 ; find the centre of gravity of 
the system. 



§ 163. Centre of Gravity of Homogeneous Bodies. — For 

the case of a single homogeneous body, the formulae have been 
already deduced in § 44. They are 

_ fxdV _ fydV fzdV 

Xo -7dv> y °- fdV z °~~ fdV> 

and for the weight of the body, 

W = wfdV, 

where x oy y oi z , are the co-ordinates of the centre of gravity of 
the body, W its weight, and w its weight per unit of volume. 

From these formulae we can readily deduce those for any 
special cases ; thus, — 

(a) For a volume referred to rectangular co-ordinate axes, 
dV — dxdydz. 

_ fffxdxdydz __ fffydxdydz __ fffzdxdydz 

X ° ~~~ Sffdxdydz' y ° ~ fffdxdydz *° ~ fffdxdydz' 

(b) For a flat plate of uniform thickness, t, the centre of grav- 
ity is in the middle layer ; hence only two co-ordinates are 
required to determine it. If it be referred to a system of rect- 
angular axes in the middle plane, dV '= tdxdy, 

x _ ffxdxdy __ ffydxdy 

ffdxdy ' ° ffdxdy ' 



214 APPLIED MECHANICS. 

The centre of gravity of such a thin plate is also called the 
centre of gravity of the plane area that constitutes the middle 
plane section ; hence — 

(c) For a plane area referred to rectangular co-ordinate axes 
in its own plane, 

_ ffxdxdy __ ffydxdy 

ffdxdy ' f/dxdy' 

(d) For a slender rod of tmiform sectional area> a, if x> y, z f 
be the co-ordinates of points on the axis (straight or curved) of 
the rod, we shall have dV — ads == a\J(dx) 2 + (dy) 2 + (dz) 2 t 



fxds 


W.+(i 


:M0* 


*° ~ fds 


SMi 


)' ♦ Hh 


r - Syds - 


W- + (l 


)' * (§)'- 


* ~ fds - 


J'Mi 


)' + (I)'- 


_/*&■. 


yV. + (l 


V ldz\ 



fds 



./*+<#♦ ey* 



(e) For a slender rod whose axis lies wholly i?i one plane, 

the centre of gravity lies, of course, in the same plane ; and if 

our co-ordinate axes be taken in this plane, we shall have s = o 

dz 

•t- = o, and also z n — o. Hence we need only two co- 

dx 



CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 215 

ordinates to determine the centre of gravity, hence dV '= ads 
— a>J{dx) 2 + {dyf. 



/yds ^ I+ (ih 

Ids 

dx 



/v/. + (|)'- 



(/) For a line, straight or curved, which lies entirely in one 
plane, we shall have, again, 



V-dT 



feeds fx V ' 
*° - Jds 



" S)h*lS 



dx 



«, = £* 

J as 



., WMih 



U>+(£h 



Whenever the body of which we wish to determine the 
centre of gravity is made up of simple figures, of which we 
already know the positions of the centres of gravity, the method 
explained in § 162 should be used, and not the formulae that 
involve integration ; i.e., taking moments about any given line 
will give us the perpendicular distance of the centre of gravity 
from that line. 

In the case of the determination of the strength and stiff- 
ness of beams, it is necessary to know the distance of a hori- 
zontal line passing through the centre of gravity of the section, 



2l6 APPLIED MECHANICS. 

from the top or the bottom of the section ; but it is of no prac- 
tical importance to know the position of the centre of gravity 
on this line. In most of the examples that follow, therefore, 
the results given are these distances. These examples should 
be worked out by the student. 

In the case of wrought-iron beams of various sections, on 
account of the thinness of the iron, a sufficiently close approxi- 
mation is often obtained by considering the cross-section as 
composed of its central lines ; the area of any given portion 
being found by multiplying the thickness of the iron by the 
corresponding length of line, the several areas being assumed 
to be concentrated in single lines. 

EXAMPLES. 

i. Straight Line AB (Fig. 127). — The centre of gravity is evidently 
at the middle of the line, as this is a point of 
symmetry. 



c 



B 



Fig. 127. 

2. Combination of Two Straight Lines. — The centre of gravity in 
each case lies on the line OO t , Figs. 128, 129, 130, and 131. 

(a) Angle- Iron of Unequal Arms (Fig. 128). — Length AB = b, 
length BC = h, area AB = A, area 
BC = B; 



BE == DE = - 1 - 



bh 

2. 




'V^ 2 ■+- h 2 

(b) Angle-Iron of Equal Ar??is (Fig. 129). — Length AB = BC 
B = b ; 

b b ,_ 

/. BE = DE = — 7= = -VJ 




CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 217 



{/) Cross of Equal Arms (Fig. 130). — AB = 00 x — h; 



.-. AC = BC = 



B 
Fig 130. 

(d) T-Iron (Fig. 131). — Area AB = A, area CE = B, length 
A e b CE — h; 

Bh 



DE = 



2(a + By 



c 

Fig. 131. 



CD 



2A + B 
2(A + B) 



h. 



3. Combination of Three Lines. — 00 \ = line passing through the 
centre of gravity. 

(a) Thin Isosceles Triangular Cell (Fig. 132). — Length AB = 
BC = «, length AC = b, area ^ = .#C 
= ^, area ^4 C = B ; 

:. DB 



\/,_? 



= tf/( 2 a - b)( 2 a + J) 
A 







2(2^ + £) 

^ + B 

2(2 A + ,9) 



V(2^~ ^)(2fl + £), 



V(2a -b)( 2 a + &). 



(b) Same in Different Fosition (Fig. 133). 



BD = DC = 




Fig. 133. 



218 



APPLIED MECHANICS. 



(c) Channel-Iron (Fig. 134). — Area of flanges = A t area of web 
= B, depth of flanges -f- ^ thickness of 
web = h; 

Ah 
-o. 



CE = ij~+B' 



Fig. 134. 



„ „ J1A + 2B 



(d) T-Beam (Fig. 135). — Area of upper flange = A lf area of 
lower flange = A 2 , area of web = B, height = h. 



CQ J L 2 A 1 + B 



2 A, + A 2 + B 



gd = h 2 A l + B 



C B 
G 



-o, 



E D F 

Fig. 135. 



2 A t + A 2 + B 

4. Co7nbination of Four Lines. — 00 1 = line passing through the 
centre of gravity. 

(a) Thin Rectangular Cell (Fig. 136). — Length AB = h; 



D A 
F E 



c b 



-o, 



'. AE = BE 



Fig. 136. 

(b) Thin Square Cell (Fig. \$i). — AB = BC = h; 

h 



BE = CE = 



A B 

F E 



Fig. 137. 



5. Circular Arcs. 

(a) Circular Arc AB (Fig. 138). — Angle A OB = 0„ radius = r. 



Use formula 




B 



fxds 
Ids*' 



fyds 

fds '' 



but use polar co-ordinates, where 
Fig. 138. ds = rdd, x — rcosO, y = rsinfl, 



CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 219 



'£■ 



r 2 I cos Ode 






jo = 



•1 

fsu 



[ dO 

sin 6d0 



(sin <9 r ) 
IT' 



■r- 



r \ dO 



(1 — cos0 t ) sin 2 -^ 



ft 



= 2r- 



(<£) Circular Arc AC (same figure). 



r sin #! 

*o = — n > Jo = O. 



(c) Quarter- Arc of Circle AB, Radius r (Fig. 139). 



cos 



x a = 



•f< 



27* 

7T 



27* 



Jo = 




(^) Semi-circumference ABC (same figure). 



2r 



Jo = O. 



6. Combination of Circles and Straight Lines. 

Barlow Rail (Fig. 140). — Two quadrants, radius r, and web, 
c , whose area = T 3 T the united area of the quadrants. 
P Let united area of quadrants = A, area of web 
= &A; \et£J? = x Q : 




TT^^o 



<V0 = 



.22 

7 J 



ft^' 



5 = ^- 



220 



APPLIED MECHANICS. 



7. Areas. 

O) T-Section (Fig. 141). — Let length AB = B, EF = b, entire 
height = H, GE =■ h. Let distance of centre 
X iX\ of gravity below AB = x*\ therefore, taking 

1 moments about AB as an axis, 

x x \BH - h(B - b)\ 

= \BH* - h(B - b)\H - -\ 
whence we can readily derive x x . 




{b) I-Section (Fig. 142). — Let AB = B, GH = b, MN = b„ 
entire height = H, BC = H — h, EH = k t ; and let x x = distance 
of centre of gravity below AB. A b 

1 _ Jq 



Hence, taking moments about AB, we have 



x,\B{H - h) + b x {h - h^ + bh Y \ 



= -(H - hy + bh 

+ b x (h - h,)\H 
whence we can deduce x t . 



i'-t 



h - h x 



G H 

Fig. 142. 



(c) Triangle (Fig. 143). — If we consider the triangle OBC as 
composed of an indefinite number of narrow strips parallel to the side 
CB, of which FLHK is one, the centre of gravity 
of each one of these strips will be on the line OD 
drawn from O to the middle point of the side 
CB ; hence the centre of gravity of the entire tri- 
angle must be on the line OD. For a similar rea- 
son, it must be on the median line CE ; hence the 
centre of gravity must be at the intersection of the median lines, and 
hence 

^ BC . OD sin ODC 
x Q = OG — \OD. Moreover, area = 





CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 22 



(d) Trapezoid (Fig. 144). 

Graphical Solution. — Bisect AB in O, and CE in D; join O 

and D ; consider the trapezoid to be divided into two triangles, ABC 

and CEB ; let g l be the centre of gravity of c p_ 

CEB, and g 2 that of ABC. Then will 

DG X = G X G 2 = 0G 2 = \0D, 

where G^ || G 2 g 2 || ^4i?. Then will G, the centre of gravity of the 
trapezoid, be on the line g^, at such a point as shall make 

G& _ ABC 
Gg 2 ~ CEB' 

But a similar reasoning to that used in the case of the triangle will show 
that it must be on the line OD ; hence it must be at the intersection of 
OD and g z g 2 . This gives a graphical construction, and from this we can 
deduce the value of OG as follows : — 

From the similarity of GG x g, and GG 2 g 2 we have 

GG, _ Gg x _ ABC _AB _B 
GG 2 ~ Gg 2 ~ BEC ~ CE = T' . 



since the triangles ABC and CEB have the same altitude : hence, from 

• GG * B , 

the proportion -^r = -r-, we have 

£^2 ~ B + j' 

but Gi G 2 = , and hence 

3 

3 3 \ B + £/ 



222 . APPLIED MECHANICS. 

(e) Parabolic Half-Segment OAB (Fig. 145). — Let OA — x ZJ 
AB = y\ \ let x , y OJ be the co-ordinates of the centre of gravity ; let 
the equation of the parabola be y 2 = ^ax : 

- 



Xq — 



J I xdxdy 2a- I x l dx 

O I/O t/ 

I dxdy 2a J o x ' l(ix 



_ jxfi 

Fig. 145. - 2 I - **« 

3 ^ z 



Jr»2aM /^^ 
I ydxdy 
o *Jo 

o t/o 






.*,» 



Area =1 / dxdy = %a?xs — \{*&x$)x x = §*,>? z . 

t/O 1/Q 



(/) Parabolic Spandril OBC (Fig. 145). — Let ^ oj JJ ; oj be co-ordi- 
nates of centre of gravity of the spandril. 



J I xdxdy 

2a$x$ */o 

Jrvi 7^ 



— To*n 



I ydxdy 

2rtM *^o _ ' 3 

T^ — f*» ~~ 4 



Area = x,y t - ^j/, = Jr,.?, 



CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 223 



(g) Circular Sector OAC (Fig. 146). — Let OA = r, A OX 
*o, y , be the co-ordinates of the centre of gravity ; 



C\jr 2 -x 2 fr f*x\3ia.0 x /Vcosfli 

/ I xdxdy -f- I 1 xdxc 

_ J-^yz — x 2 *Jrcos9 x ^--rtang! *^o 



xdxdy 



ir(2rd s ) 



%r 



sine/! 



Fig. 146. 



Area = \r(2rB x ) = r 2 Q x . 

K Second Solution. 

Consider the sector to be made up of an indefinite number of narrow 
rings ; let p be the variable radius, and dp the thickness ; 

.'. Elementary area = 2pO l dp, 

and centre of gravity of this elementary area is on OX, at a distance 

. sin 6 T 

from O equal to p— — [see Example 5 (3)]; 

J \ pS ^Tr\ \ 2 ? 6ldp \ 2 sin 0, J r p 2 dp 



i> 



dp 



26, j pdp 



_ 2 Sin( ^x 



(k) Circular Half-Segment ABX (Fig. 146). 

Jf^ 2 --^ 2 /V /V J 
I xdxdy I .#VV 2 — ^ 2 ^r 
o J r cos fl t _ Jr cos X 

~~ Sector minus triangle ~~ \r 2 d x — %r 2 sin X cos 0, 



sin^ 



J^V^ 2 --* 2 /v 
o *>Vcos6> x 

\r*{B x — sin# x cos0 x ) 



3 0, — sin 0, cos fl, 



' 4 sin 2 -|0 x — sin 2 <9 X cos 6 t 
3 ' 0/ - sin 0, cos X 



224 AT PLIED MECHANICS. 

§ 164. Pappus's Theorems. — The following two theorems 
serve often to simplify the determination of the centres of 
gravity of lines and areas. They are as follows : — 

Theorem I. — If a plane curve lies wholly on one side of a 
straight line in its own plane, and, revolving about that line, 
generates thereby a surface of revolution, the area of the sur- 
face is equal to the product of the length of the revolving line, 
and of the path described by its centre of gravity. 

Proof. — Let the curve lie in the xy plane, and let the axis 
of y be the line about which it revolves. We have, from what 

precedes, § 163 (*), x Q — — r- ; 

.*. x Q fds = fxds, 

where x Q equals the perpendicular distance of the centre of 
gravity of the curve from OY, ds = elementary arc, 

.-. 2irx Q fds = f(2irx)ds; 

or, reversing the equation, 

f(27TX)ds = (2-x )s. 

But f(2-x)ds = surface described in one revolution, while s = 
length of arc, and 2-x Q c= path described by the centre of 
gravity in one revolution. Hence follows the proposition. 

Theorem II. — If a plane area lying wholly on the same 
side of a straight line in its own plane revolves about that line, 
and thereby generates a solid of revolution, the volume of the 
solid thus generated is equal to the product of the revolving 
area, and of the path described by the centre of gravity of the 
plane area during the revolution. 



PAPPUS'S THEOREMS. 22$ 

Proof. — Let the area lie in the xy plane, and let the axis 
OY be the axis of revolution. We then have, from what has 
preceded, if x = perpendicular distance of the centre of gravity 
of the plane area from OY, the equation, § 163 {b), 

ffxdxdy 



x Q = 



Hence 



ffdxdy 



or 



x ffdxdy — ffxdxdy ; 

.*. (27rx ) ffdxdy = ff(27rx)dxdy 

ff(27rx)dxdy = 2irx Q fdxdy. 

But ff(2-rrx)dxdy = volume described in one revolution, and 
2irx Q = path described by the centre of gravity in one revolu- 
tion. Hence follows the proposition. 

The same propositions hold true for any part of a revolution, 
as well as for an entire revolution, since we might have multi- 
plied through by the circular measure 6, instead of by 2n. 

It is evident that the first of these two theorems may be 
used to determine the centre of gravity of a line, when the 
length of the line, and the surface described by revolving it 
about the axis, are known ; and so also that the second theorem 
may be used to determine the centre of gravity of a plane area 
whenever the area is known, and also the volume described by 
revolving it around the axis. 

EXAMPLES. 

i. Circular Arc AC (Fig. 138). — Length of arc = s = 2r$, sur- 
face of zone described by revolving it about OY '= circumference of a 
great circle multiplied by the altitude = (2-n-r) {2rsm6 1 ); 

.*. (2ttx ) (2r6j) =(27rr)(2rsm6 1 ) .'. x o 1 = rsinO s 

sin0! 
/. x = r^ — • 



226 APPLIED MECHANICS. 



2. Semicircular Arc (Fig. 139). — Length of arc = 7rr, surface of 
sphere described = 4-n-r 2 ; 

. . 2r 

.'. 27rx (7rr) = 4-n-r 2 .'. x Q = — 

IT 

3. Trapezoid (Fig. 147). — Let AD = b, BC = b ; let it revolve 
around AD : it generates two cones and a cylinder. 

Area of trapezoid = BG, 

tt(GB) 2 

B Volume = — -{AG + HD) + tt(GB) 2 . BC 

o 

E tt(GB) 2 

c =-±-j±(AG + HD+sBC) 

it(GB) 2 

Fig - J 47. = -* J - (AD + BC -t BC) 

... (MO p*±*£) ^ = Hgb }1] {AD + BC) + BC{ 






GB( BC \ GBl b \ 

*« =^{^ADTBC)=TV + B^) = XL 



FE( 



+ b) 



4. Circular Sector ACO (Fig. 146). — Area of sector = r 2 6„ 

volume described = -|r(surface of zone) = \r(2-r) (2r sin #,) = 

f 7rr 3 sin 0, ; 

sin r 
.-. (2Trx )(r 2 6 l ) = %irr*sinO x .*. x = f*-- 



0, 

§ 165. Centre of Gravity of Solid Bodies. — The general 
formulae furnish, in most cases, a very complicated solution, and 
hence we generally have recourse to some simpler method. A 
few examples will be given in this and the next section. 



CENTRE OF GRAVITY OF SYMMETRICAL BODIES. 22 J 




Tetrahedron ABCD (Fig. 148). — The plane ABE, containing the 
edge AB and the middle point E of the edge CD, bisects all lines 
drawn parallel to CD, and terminating in the faces 
ABD and ABC : hence a similar reasoning to that 
used in the case of the triangle will show that the cen- 
tre of gravity of the pyramid must be in th£ plane 
ABE ; in the same way it may be shown that it must 
lie in the plane ACE. Hence it must lie in their 
intersection, or in the line AG joining the vertex A 
with the centre of gravity (intersection of the medians) 
of the opposite face. In the same way it can be shown that the centre 
of gravity of the triangular pyramid must lie in the line drawn from 
the vertex B to the centre of gravity of the face A CD. Hence the 
centre of gravity of the tetrahedron will be found on the line AG at 
a distance from G equal to \A G. 

§ 166. Centre of Gravity of Bodies which are Symmet- 
rical with Respect to an Axis. — Such solids may be gener- 
ated by the motion of a plane figure, as ABCD 
(Fig. 149), of variable dimensions, and of any 
form whose centre G remains upon the axis 
OX ; its plane being always perpendicular to 
OX, and its variable area X being a function 
of x, its distance from the origin. 

Here the centre of gravity will evidently 
lie on the axis OX, and the elementary vol- 
ume will be the volume of a thin plate whose area is X and 
thickness Ax ; hence the elementary volume will be XAx. 




Fig. 149. 



or 



Take moments about OY, and we shall have 

x Q fXdx = fXxdx and Volume = /Xdx, 
fXxdx 



x = 



/Xdx 



V = fXdx. 



228 



APPLIED MECHANICS. 



EXAMPLES. 

x 2 y 2 z 2 

I. Ellipsoid — + 7- -\ — - = i (Fig. 150). — Find centre of gravity 

a. D C 

z of the half to the right of the x plane. Let OK 

= x. Now if, in the equation of the ellipsoid, 

x x 2 z 2 

we make y = o, we have — + — = 1 ; 




z = ~>Ja 2 — x 2 , 
a 



where z = jE'A'. 



.x 2 _y* 



Make 2 = o in the equation of the ellipsoid, and — + ^ = 1 ; 

<*/ 

.*. y = -\a 2 — x 2 . 

where jy = KG; 

.-. EK = -S/a 2 - x 2 , KG = -\/a 2 - x 2 , 



are the semi-axes' of the variable ellipse EGFH, which, by moving along 
OX, generates the ellipsoid. Hence 



hence 



Area EGFH = tt(EK . GK) = — («' - x 2 ) = X; 



irbc 

Elementary volume = — j (a 2 — x 2 )kx 

— I (a 2 x — x*)dx < > 

a 2 J V ' \ 2 4 J, 



l«. 



— I (a 2 - x 2 )dx \a 2 x \ 

a 2 Jo ( 3 'o 

F = — / (a 2 - * 2 K* ^ = \ieabc. 
2. Hemisphere. — Make a = b — c, and x Q = §0, F= -f^tf 3 . 



CENTRE OF GRAVITY OF SYMMETRICAL BODIES. 229 



If the section X were oblique to OX, making an angle 
with it, the elementary volume would not be Xdx, but Xdx sin 0, 
and we should have 



sin O/Xxdx fXxdx 
sin OfXdx == /Xdx 



and V — sin 6 J Xdx. 



3. Oblique Cone (Fig. 151). — Let OA == h; let area of base be 
A, and let the angle made by OX with the base be 6 ; 



■ X x ° ■ y 
•■ A~T> ■• X 


A 
h 2 




Z ^-7T-^ 


~ A C k J ~ hz " 4 ''* / 
frJo 3 Y 


Fig. 151. 


V = sin 0/X/* = —sin j x 2 


s& = \Ahsm6. 



4. Truncated Cone (Fig. 151). — Let height of entire cone be 
h — OA ; let height of portion cut off be h t ; 



Xq — 



feJiL 4 



— / x 2 dx 
KJh x 



fc - h* *& - h/ 



A „ C h ■ Ah sin $ / hj\ 



230 APPLIED MECHANICS. 



CHAPTER VI. 

STRENGTH OF MATERIALS. 

§ 167. Stress, Strain, and Modulus of Elasticity. — When 
a body is subjected to the action of external forces, if we 
imagine a plane section dividing the body into two parts, the 
force with which one part of the body acts upon the other 
at this plane is called the stress on the plane ; and, in order 
to know it completely, we must know its distribution and its 
direction at each point of the plane. If we consider a small 
area lying in this plane, including the point O, and represent 
the stress on this area by p } whereas the area itself is repre- 
sented by a, then will the limit of -c- as a approaches zero be the 

a 

intensity of the stress on the plane under consideration at the 
point O. 

When a body is subjected to the action of external forces, 
and, in consequence of this, undergoes a change of form, it 
will be found that lines drawn within the body are changed, by 
the action of these external forces, in length, in direction, or 
in both ; and the entire change of form of the body may be 
correctly described by describing a sufficient number of these 
changes. 

If we join two points, A and £, of a body before the 
external forces are applied, and find, that, after the application 
of the external forces, the line joining the same two points of 
the body has undergone a change of length A(AB), then is the 



STRESS, STRAIN, AND MODULUS OF ELASTICITY. 23 1 



limit of the ratio — — — - as AB approaches zero called the 

strain of the body at the point A in the direction AB. 

If AB -f- A(AB) > AB, the strain is one of tension. 

If AB + A(AB) < AB, the strain is one of compression. 

Suppose a straight rod of uniform section A to be subjected 
to a pull P in the direction of its length, and that this pull is 
uniformly distributed over the cross-section : then will the in- 
tensity of the stress on the cross-section be 

If P be measured in pounds, and A in square inches, then will 
/ be measured in pounds per square inch. 

If the length of the rod before the load is applied be /, 
and its length after the load is applied be / -f- e, then is e the 
elongation of the rod; and if this elongation is uniform through- 
out the length of the rod, then is - the elongation of the rod 

per unit of length, or the strain. 

Hence, if a represent the strain due to the stress / per 
unit of area, we shall have 

e 

a = -. 

/ 

The Modulus of Elasticity is the quotient obtained by 
dividing the stress per unit of area by the strain, or 

a 

and this is expressed in units of weight per unit of area, as in 
pounds per square inch. 



232 APPLIED MECHANICS. 

The modulus of elasticity is sometimes defined as the 
weight that would be required to stretch a rod one square inch 
in section to' double its length, if Hooke's law, "The stress is 
proportional to the strain," held up to that point, and the rod 
did not break. Of course the modulus of elasticity is a con- 
stant for any given material just as far as, and no farther than, 
Hooke's law holds. 

EXAMPLES. 

i. A wrought-iron rod 10 feet long and i inch in diameter is loaded 
in the direction of its length with 8000 lbs. ; find (1) the intensity of 
the stress, (2) the elongation of the rod ; assuming the modulus of the 
iron to be 28000000 lbs. per square inch. 

2. What would be the elongation of a similar rod of cast-iron 
under the same load, assuming the modulus of elasticity of cast-iron to 
be 1 7000000 lbs. per square inch ? 

3. Given a steel bar, area of section being 4 square inches, the 
length of a certain portion under a load of 25000 lbs. being 10 feet, 
and its length under a load of 1 00000 lbs. being io' o".o75 ; find the 
modulus of elasticity of the material. 

4. What load will be required to stretch the rod in the first example 
T Vinch? 

§ 168. Resistance to Stretching and Tearing. — The most- 
used criterion of safety against injury for a loaded piece is, 
that the greatest intensity of the stress to which any part of it 
is subjected shall nowhere exceed a certain fixed amount, called 
the working-strength of the material ; this working-strength 
being a certain fraction of the breaking-strength determined 
by practical considerations. 

The more correct but less used criterion is, that the great- 
est strain in any part of the structure shall nowhere exceed 
the working-strain ; the greatest allowable amount of strain 
being a fixed quantity determined by practical considerations. 



RESISTANCE TO STRETCHING AND TEARING. 233 

This is equivalent to limiting the allowable elongation or 
compression to a certain fraction of its length, or the deflection 
of a beam to a certain fraction of the span. 

If the stress on a plane surface be uniformly distributed, 
its resultant will evidently act at the centre of gravity of the 
surface, as has been already shown in § 42 to be the case with 
any uniformly distributed force. 

If a straight rod of uniform section and material be sub- 
jected to a pull in the direction of its length, and if the result- 
ant of the pull acts along a line passing through the centres 
of gravity of the sections of the rod, it is assumed in practice 
that the stress is uniformly distributed throughout the rod, and 
hence that for any section we shall obtain the stress per square 
inch by dividing the total pull by the number of square inches 
in the section. 

If, on the other hand, the resultant of the pull does not act 
through the centres of gravity of the sections, the pull is not 
uniformly distributed ; and while 



will express the mean stress per square inch, the actual inten- 
sity of the stress will vary at different points of the section, 

p 

being greater than — at some points and less at others. How 

to determine its greatest intensity in such cases will be shown 
later. 

With good workmanship and well-fitting joints, the first 
case, or that of a uniformly distributed stress, can be practi- 
cally realized ; but with ill-fitting joints or poor workmanship, 
or with a material that is not homogeneous, the resultant of 
the pull is liable to be thrown to one side of the line passing 
through the centres of gravity of the sections, and thus there 



234 APPLIED MECHANICS. 

is set up a bending-action in addition to the direct tension, and 
therefore an unevenly distributed stress. 

It is of the greatest importance in practice to take cogni- 
zance of any such irregularities, and determine the greatest 
intensity of the stress to which the piece is subjected: though 
it is too often taken account of merely by means of a factor of 
safety ; in other words, by guess. 

Leaving, then, this latter case until we have studied the 
stresses due to bending, we will confine ourselves to the case 
of the uniformly distributed stress. 

If the total pull on the rod in the direction of its length 

be P, and the area of its cross-section A, we shall have, for the 

intensity of the pull, 

P 

* = A 

On the other hand, if the working-strength of the material 

per unit of area be f t we shall have, for the greatest admissible 

load to be applied, 

P = /A. 

If f be the working-strength of the material per square 
inch, and E the modulus of elasticity, then is the greatest 
admissible strain equal to 

Thus, assuming 12000 lbs. per square inch as the working 
tensile strength of wrought-iron, and 28000000 lbs. per square 
inch as its modulus of elasticity, its working-strain would be 

12000 * 



28000000 7000 



Hence the greatest safe elongation of the bar would be 
Y^ S7 of its length. Hence a rod 10 feet long could safely be 
stretched Y j^ of a foot = 0.05 14". 



VALUES OF BREAKING AND WORKING STRENGTH. 2$$ 

§ 169. Approximate Values of Breaking and Working 
Strength, and of Modulus of Elasticity. — In a later part of 
this book the attempt will be made to give an account of the 
experiments that have been made to determine the strength 
and elasticity of the materials ordinarily used in construction, 
in such a way as to enable the student to decide for himself, in 
any special case, upon the proper values of the constants that 
he ought to use. 

For the present, however, the following will be given as a 
rough approximation to some of these quantities, which we may 
make use of in our work until we reach the above-mentioned 
account. 

(a) Cast-Iron. 

Breaking tensile strength per square inch, of common quali- 
ties, 14000 to 20000 lbs. ; of gun iron, 30000 to 33000 lbs. 

Modulus of elasticity for tension and for compression, about 
17000000 lbs. per square inch. 

(b) Wronght-Iron. 

Breaking tensile strength per square inch, from 40000 to 
60000 lbs. 

Modulus of elasticity for tension and for compression, about 
28000000. 

(c) Mild Steel. 

Breaking tensile strength per square inch, 55000 to 70000 
lbs. 

Modulus of elasticity for tension and for compression, from 
28000000 to 30000000 lbs. per square inch. 

id) Wood. 

Breaking compressive strength per square inch : — 

Oak, green 3000 lbs. 

Oak, dry 3000 to 6000 lbs. 

Yellow pine, green 3000 to 4000 lbs. 

Yellow pine, dry 4000 to 7000 lbs. 



236 APPLIED MECHANICS. 



Modulus of elasticity for compression (average values) : — 

Oak 1300000 lbs. per square inch. 

Yellow pine 1600000 lbs. per square inch. 

§ 170. Sudden Application of the Load. — If a wrought- 
iron rod 10 feet long and 1 square inch in section be loaded 
with 12000 pounds in the direction of its length, and if the 
modulus of elasticity of the iron be 28000000, it will stretch 
0.05 14" provided the load be gradually applied: thus, the rod 
begins to stretch as soon as a small load is applied ; and, as the 
load gradually increases, the stretch increases, until it reaches 
0.05 14". 

If, on the other hand, the load of 12000 lbs. be suddenly 
applied (i.e., put on all at once) without being allowed to fall 
through any height beforehand, it would cause a greater stretch 
at first, the rod undergoing a series of oscillations, finally 
settling down to an elongation of 0.05 14". 

To ascertain what suddenly applied load will produce at 
most the elongation 0.05 14", observe, that, in the case of the 
gradually applied load, we have a load gradually increasing from 

o to 12000 lbs. 

Its mean value is, therefore, J(i2000) = 6000 lbs. ; and this 
force descends through a distance of 

0.05 14". 

Hence the amount of mechanical work done on the rod by the 
gradually applied load in producing this elongation is 

(6000) (0.0514) = 308.4 inch-lbs. 

Hence, if we are to perform upon the rod 308.4 inch-lbs. of 
work with a constant force, and if the stretch is to be 0.05 14", 
the magnitude of the force must be 

-^d_ . 6000 lbs. 
0.0514 



RESILIENCE OF A TENSION-BAR. 2^7 

Hence a suddenly applied load will produce double the strain 
that would be produced by the same load gradually applied ; 
and, moreover, a suddenly applied load should be only half as 
great as one gradually applied if it is to produce the same 
strain. 

§ 171. Resilience of a Tension-Bar. — The resilience of a 
tension-rod is the mechanical work done in stretching it to the 
same amount that it would stretch under the greatest allowable 
gradually applied load, and is found by multiplying the greatest 
allowable load by half the corresponding elongation. 

Thus, suppose a load of 100 lbs. to be dropped upon the 
rod described above in such a way as to cause an elongation not 
greater than 0.05 14", it would be necessary to drop it from a 
height not greater than 5.0%". 

EXAMPLES. 

1. A wrought- iron rod is 12 feet long and t inch in diameter, and 
is loaded in the direction of its length ; the working- strength of the 
iron being 12000 lbs. per square inch, and the modulus of elasticity 
28000000 lbs. per square inch. 

Find the working-strain. 
Find the working-load. 
Find the working-elongation. 
Find the working-resilience. 

From what height can a 50-pound weight be dropped so as to produce 
tension, without stretching it more than the working- elongation? 

2. Do the same for a cast-iron rod, where the working-strength is 
5000 pounds per square inch, and the modulus of elasticity 17000000; 
the dimensions of the rod being the same. 

§ 172. Results of Wohler's Experiments on Tensile 

Strength. — According to the experiments of Wohler, of which 
an account will be given later, the breaking-strength of a piece 



238 APPLIED MECHANICS. 

depends, not only on whether the load is gradually or suddenly 
applied, but also on the extreme variations of load that the 
piece is called upon to undergo, and the number of changes to 
which it is to be submitted during its life. 

For a piece which is always in tension, he determines the 
following two constants ; viz., t, the carrying-strength per square 
inch, or the greatest quiescent stress that the piece will bear, 
and ?/, the primitive safe strength, or the greatest stress per 
square inch of which the piece will bear an indefinite number 
of repetitions, the stress being entirely removed in the inter- 
vals. 

This primitive safe strength, u t is used as the breaking- 
strength when the stress varies from o to u every time. Then, 
by means of Launhardt's formula, we are able to determine the 
ultimate strength per square inch for any different limits of 
stress, as for a piece that is to be alternately subjected to 80000 
and 6000 pounds. 

Thus, for Phoenix Company's axle iron, Wohler finds 

/ = 3290 kil. per sq. cent. = 46800 lbs. per sq. in., 
u = 2100 kil. per sq. cent. = 30000 lbs. per sq. in. 

Launhardt's formula for the ultimate strength per unit of area 
is 

( t — u least stress ) 

a = u I 1 -| / . 

( u greatest stress) 

Hence, with these values of t and u, we should have, for the 
ultimate strength per square inch, 

( 1 least stress ) 

a — 2100^' 1 -f ->ku. per sq. cent., 

( 2 greatest stress) l 



or 



( 1 least stress ) 

30000 { 1 + > lbs. per sq. in. 

{ 2 greatest stress \ 



WOHLER'S EXPERIMENTS ON TENSILE STRENGTH. 239 

Thus, if least stress =s 6000, and greatest =a 80000, we should 
have 

a = 30000J1 4- J • g 6 o? = S 0000 ! 1 + s 3 oS = 3 II2 5i 

if least stress = 60000, and greatest = 80000, 

a = 30000 J 1 -f J. %\ = 30000J1 + f J = 41250; 

if least stress = greatest stress = 80000, 

a = 30000J1 4- \\ = 45000 = carrying-strength. 

Hence, instead of using, as breaking-strength per square inch 
in all cases, 45000, we should use a set of values varying from 
45000 down to 30000, according to the variation of stress which 
the piece is to undergo. 

For working-strength, Weyrauch divides this by 3 : thus 
obtaining, for working-strength per square inch, 

, ( , 1 least stress ) „ 

b = 10000 H > lbs. per sq. in. ; 

( 2 greatest stress ) 

for Krupp's cast-steel, 

/ = 7340 kil. per sq. cent. = 104400 lbs. per sq. in., 

u = 3300 kil. per sq. cent. = 46900 lbs. per sq. in. ; 



or 



( o least stress ) 

a = 3300 < 1 4- — } kil. per sq. cent., 

«" ( 11 greatest stress j l H ' 

( o least stress ) 

a = 46900 < 1 4- — — wbs. per sq. in., 

y { 11 greatest stressf 



- 15633I 



o least .stress ) 

1 + 7~t . . . 1 lbs. per sq. in. 

1 1 greatest stress j 



EXAMPLES. 

Find the breaking-strength per square inch for a wrought-iron tension- 
rod. 

1. Extreme loads are 75000 and 6000 lbs. 

2. Extreme loads are 120000 and 100000 lbs. 

3. Extreme loads are 300000 and 10000 lbs. 
Find the safe section for the rod in each case. 



24O APPLIED MECHANICS. 

§173. Suspension-Rod of Uniform Strength. — In the 
case of a long suspension-rod, the weight of the rod itself some- 
times becomes an important item. The upper section must, of 
course, be large enough to bear the weight that is hung from 
the rod plus the weight of the rod itself; but it is sometimes 
desirable to diminish the sections as they descend. This is often 
accomplished in mines by making the rod in sections, each section 
being calculated to bear the weight below it plus its own weight. 
Were the sections gradually diminished, so that each section 
would be just large enough to support the weight below it, we 
should, of course, have a curvilinear form ; and the equation of 
this curve could be found as follows, or, rather, the area of any 
section at a distance from the bottom of the rod. 
Let W — weight hung at O (Fig. 152), 

Let w = weight per unit of volume of 

the rod, 

Let x = distance AO, 

Let S = area of section A, 

Let x -f- dx = distance BO, 



- — s rtr 

Let 5 + dS = area of section at B f 

1 
1 



Let f = working-strength of the mate- 

i rial per square inch. 

i°. The section at O must be just large enough 



to sustain the load W t 



W 

Fig. 152. •*• ^o == r ' 

2°. The area in dS must be just enough to sustain the 
weight of the portion of the rod between A and B. 
The weight of this portion is wSdx ; 

wSdx 



d$ = 



7 



dS w , • W 

— = — dx :. log, S = -tx + a constant. 



CYLINDERS SUBJECTED TO INTERNAL PRESSURE. 24I 



w 

When x = o, 5 = -j ; 



, w L 

*. \og e —F = the constant 


,\ \og e S — log, 


5 = 

— : pJ 





k')-r 



This gives us the means of determining the area at any dis- 
tance x from O. 

EXAMPLES. 

1. A wrought-iron tensiqn-rod 200 feet long is to sustain a load of 
2000 lbs. with a factor of safety of 4, and is to be made in 4 sections, 
each 50 feet long; find the diameter of each section, the weight of the 
wrought iron being 480 lbs. per cubic foot. 

2. Find the diameter needed if the rod were made of uniform 
section, also the weight of the extra iron necessary to use in this case. 

3. Find the equation of the longitudinal section of the rod, assum- 
ing a square cross-section, if it were one of uniform strength, instead of 
being made in 4 sections. 

§ 174. Thin Hollow Cylinders subjected to an Internal 
Normal Pressure. — Let/ denote the uniform intensity of the 
pressure exerted by a fluid which is confined within a hollow 
cylinder of radius r and of thickness t (Fig. 153), e 

the thickness being small compared with the radius, /j n\ 

Ills. o B\\ 

Let us consider a unit of length of the cylinder, and c (r | D 

let us also consider the forces acting on the upper ^^-^y 
half -ring CED. F * G - *53- 

The total upward force acting on this' half-ring, in conse- 
quence of the internal normal pressure, will be the same as 
that acting on a section of the cylinder made by a plane pass- 
ing through its axis, and the diameter CD. The area of this 



APPLIED MECHANICS. 



section will be 2r X i = 2r: hence the total upward force will be 
2rX/ = 2//y and the tendency of this upward force is to cause 
the cylinder to give way at A and B } the upper part separating 
from the lower. 

This tendency is resisted by the tension in the metal at the 
sections AC and BD ; hence at each of these sections, there has 
to be resisted a tensile stress equal to \{2pr) = pr. This stress 
is really not distributed uniformly throughout the cross-section 
of the metal ; but, inasmuch as the metal is thin, no serious 
error will be made if it be accounted as distributed uniformly. 
The area of each section, however, is t X i=-t; therefore, if 
Zdenotethe intensity of the tension in the metal in a tangential 
direction (i.e., the intensity of the hoop tension), we shall have 



T = K 

t 

Hence, to insure safety, T must not be greater than /, the 
working-strength of the material for tension ; hence, putting 

f= p - 
*' 

we shall have 

-'i 

as the proper thickness, when/ = normal pressure per square 
inch, and radius = r. 

The above are the formulae in common use for the deter- 
mination of the thickness of the shell of a steam-boiler; for in 
that case the steam-pressure is so great that the tension 
induced by any shocks that are likely to occur, or by the weight 
of the boiler, is very small in comparison with that induced 
by the steam-pressure. On the other hand, in the case of an 
ordinary water-pipe, the reverse is the case. 



RESISTANCE TO DIRECT COMPRESSION. 243 

To provide for this case, Weisbach directs us to add to the 
thickness we should obtain by the above formulae, a constant 
minimum thickness. 

The following are his formulae, d being the diameter in 
inches, p the internal normal pressure in pounds per square 
inch, and / the thickness in inches. For tubes made of 

Sheet-iron t = 0.00086/*/ +0.12 

Cast-iron . . . t = 0.00238/;/ -f 0.34 

Copper t = 0.00148/;/+ 0.16 

Lead t = 0.00507/;/ 4- 0.21 

Zinc / = 0.00242 pd 4- 0.16 

Wood / = 0.03230/;/ 4- 1 .07 

Natural stone t == 0.03690/;/ 4 1.18 

Artificial stone ....... t = 0.05380/;/ 4- 1.58 

§ 175. Resistance to Direct Compression. — When a piece 
is subjected to compression, the distribution of the compressive 
stress on any cross-section depends, first, upon whether the 
resultant of the pressure acts along the line containing the cen- 
tres of gravity of the sections, and, secondly, upon the dimen- 
sions of the piece ; thus determining whether it will bend or 
not. 

In the case of an eccentric load, or of a piece of such length 
that it yields by bending, the stress is not uniformly distributed ; 
and, in order to proportion the piece, we must determine the 
greatest intensity of the stress upon it, and so proportion it 
that this shall be kept within the working-strength of the ma- 
terial for compression. 

Either of these cases is not a case of direct compres- 
sion. 

In the case of direct compression (i.e., where the stress over 
each section is uniformly distributed), the intensity of the stress 
is found by dividing the total compression by the area of the 



244 APPLIED MECHANICS. 

section ; so that, if P be the total compression, and A the area 
of the section, and/ the intensity of the compressive stress, 

>--f 

On the other hand, if f is the compressive working-strength of 
the material per square inch, and A the area of the section in 
square inches, then the greatest allowable load on the piece 
subjected to compression is 

P = fA. 

The same remarks as were made in regard to a suddenly 
applied load and resilience, in the case of direct tension, apply 
in the case of direct compression. 

§ 176. Results of Wohler's Experiments on Compressive 
Strength. — Wohler also made experiments in regard to pieces 
subjected to alternate tension and compression, taking, in the 
experiments themselves, the case where the metal is subjected 
to alternate tensions and compressions of equal amount. 

The greatest stress of which the piece would bear an indefi- 
nite number of changes under these conditions, is called the 
vibration safe strength, and is denoted by s. 

Weyrauch deduces a formula similar to that of Launhardt 
for the greatest allowable stress per unit of area on the piece 
when it is subjected to alternate tensions anal compressions of 
different amounts. 

Thus, for Phoenix Company's axle iron, Wohler deduces 

/ = 3290 kil. per sq. cent. = 46800 lbs. per sq. in., 
u = 2100 kil. per sq. cent. = 30000 lbs. persq. in., 
s = 1 1 70 kil. per sq. cent. = 16600 lbs. per sq. in. 



EXPERIMENTS ON COMPRESSIVE STRENGTH. 245 

Weyrauch's formula for the ultimate strength per unit of 
area is 

( u — s least maximum stress ) 

a = u\ 1 — 7 — : : 7 \\ 

{ u greatest maximum stress ) 

and, with these values of u and s, it gives 

least maximum stress 



a = 2ioo< 1 
or 



!i least maximum stress ) 
1 — — : = : t kil. per sq. cent., 
2 greatest maximum stress ) r ^ 

{1 least maximum stress ) 
1 — — — : > lbs. per sq. in. 
2 greatest maximum stress ) r * 



With a factor of safety of 3, we should have, for the greatest 
admissible stress per square inch, 

!i least maximum stress ) 
1 — — : = ~ 1 lbs. 
2 greatest maximum stress ) 

For Krupp's cast-steel, 

/ = 7340 kil. per sq. cent. = 104400 lbs. per sq. in., 

u — 3300 kil. per sq. cent. = 46900 lbs. per sq. in. approximately, 

s = 2050 kil. per sq. cent. = 29150 lbs. per sq. in. approximately. 

We have, therefore, for the breaking-strength per unit of 
area, according to Weyrauch's formula, 

( 5 least maximum stress ) 

a = 3300 < 1 — — — — : Vkil. per sq. cent., 

( IJ greatest maximum stress) r ^ 

or 

( 5 least maximum stress ) 

a = 46900^ 1 — — — : Ubs. per sq. in. : 

( IJ greatest maximum stress j ^ ^ 



246 APPLIED MECHANICS. 

and, using a factor of safety of 3, we have, for the greatest admis- 
sible stress per square inch, 

( e least maximum stress ) 

° = I 5^S°{ I ~ ; T7 r~. : : Mbs. per sq. in. 

( 11 greatest maximum stress J r ^ 

The principles respecting an eccentric compressive load, and 
those respecting the giving-way of long columns so far as they 
are known, can only be treated after we have studied the resist- 
ance of beams to bending; hence this subject will be deferred 
until that time. 

EXAMPLES. 

Find the proper working and breaking strength per square inch to 
be used for a wrought-iron rod, the extreme stresses being — 

1. 80000 lbs. tension and 6000 lbs. compression. 

2. 100000 lbs. tension and 100000 lbs. compression. 

3. 70000 lbs. tension and 60000 lbs. compression. 
Do the same for a steel rod. 

§ 177. Resistance to Shearing. — One of the principal cases 
where the resistance to shearing comes into practical use is 
that where the members of a structure, which are themselves 
subjected to direct tension or compression or bending, are united 
by such pieces as bolts, rivets, pins, or keys, which are sub- 
jected to shearing. Sometimes the shearing is combined with 
tension or with bending ; and whenever this is the case, it is 
necessary to take account of this fact in designing the pieces. 
It is important that the pins, keys, etc., should be equally 
strong with the pieces they connect. 

Probably one of the most important modes of connection is 
by means of rivets. In order that there may be only a shearing 
action, without any bending of the rivets, the latter must fit 
very tightly. The manner in which the riveting is done will 
necessarily affect very essentially the strength of the joints ; 



RESISTANCE TO SHEARING. 



247 



hence the only way to discuss fully the strength of riveted 
joints is to take into account the manner of effecting the rivet- 
ing, and hence the results of experiments. These will be 
spoken of later; but the ordinary theories by which the strength 
and proportions of riveted joints are determined will be given, 
which theories are necessary also in discussing the results of 
experiments thereon. 

The principle on which the theory is based is that of making 
the resistance of the joint to yielding equal in all the ways in 
which it is possible for it to yield. 



A single-riveted lap-joint is one 
with a single row of rivets, as 
shown in Fig. 154. 



A single-riveted butt-joint with 
one covering plate is shown in 
Fig. 155. 



A single-riveted butt-joint with 
two covering plates is shown in 
Fig. 156. 



JMG. 154. 




Fig. 155. 



Fig. i-,6. 



248 



APPLIED MECHANICS. 




Fig. 157. 




Fig. 158. 



A double-riveted lap-joint with 
the rivets staggered is shown in 
Fig. 157; one with chain riveting, 
in Fig. 158. 



Taking the case of the single-riveted lap-joint shown in Fig. 
154, it may yield in one of four ways : — 




i°. By the crushing of the plate 
in front of the rivet (Fig. 159). 



Fig. 160. 



2°. By the shearing of the rivet 
(Fig. 160). 




RESISTANCE TO SHEARING. 



249 



3 . By the tearing of the plate 
between the rivet-holes (Fig. 161). 




4 . By the rivet breaking 
through the plate (Fig. 162). 



Fig. 161. 



JL 



Let us call fig. 162. 

d the diameter of a rivet. 

c the pitch of the rivets ; i.e., their distance apart from 

centre to centre. 
t the thickness of the plate. 

/the lap of the plate ; i.e., the distance from the outer edge 
of a rivet-hole to the outer edge of the plate. 
f t the ultimate tensile strength of the iron. 
f s the ultimate shearing-strength of the rivet iron. 
f c the ultimate crushing-strength of the iron. 
We shall then have — 

i°. Resistance of plate in front of rivet to crushing =zfjd. 

2°. Resistance of one rivet to shearing = fs\—j- 

3 . Resistance of plate between two rivet-holes to tearing 
= f t t(c-d). 

tl 2 

4 Resistance of plate to being broken through = a — , where 

d 

a is a constant depending on the material. This may be taken 
as empirical for the present. 

An average value of this constant, as given by Robert Wil- 
son, is 100000 lbs., where all the dimensions are measured in 



250 APPLIED MECHANICS. 

inches. Assuming that we know the thickness of the plate to 
start with, we obtain, by equating the first two resistances, 

4 t /, 

which determines the diameter of the rivet. 
Equating 3 and 2°, we obtain 

f t t{c ~ d) =fs— ... c = d + J ~ 4 — , 

which gives the pitch of the rivets in terms of the diameter of 
the rivet, and the thickness of the plate. 
Equating, next, 4 and i°, we have 

«?-/•* ••• '-.*& 

which gives the lap of the plate. 

A similar method of reasoning would enable us to determine 
the corresponding quantities in the cases of double-riveted 
joints, etc. 

The above is the ordinary theory of riveted joints : it as- 
sumes that the joint should be made equally strong against 
giving way by any method in which it is possible for it to give 
way. There are a number of practical considerations which 
modify more or less the results of this theory, and which can 
only be determined experimentally. A fuller account of this 
subject from an experimental point of view will be given later. 

§ 178. Intensity of Stress. — Whenever the stress over a 
plane area is uniformly distributed, we obtain its intensity at 
each point by dividing the total stress by the area over which it 
acts, thus obtaining the amount per unit of area. When, how- 
ever, the stress is not uniformly distributed, or when its inten- 



INTENSITY OF STRESS. 



251 



sity varies at different points, we must adopt a somewhat differ- 
ent definition of its intensity at a given point. In that case, if 
we assume a small area containing that point, and divide the 
stress which acts on that area by the area, we shall have, in the 
quotient, an approximation to the intensity required, which will 
approach nearer and nearer to the true value of the intensity at 
that point, the smaller the area is taken. 

Hence the intensity of a variable stress at a given point is, — 

The limit of the ratio of the stress acting 011 a small area 
containing that point, to the area, as the latter grows smaller and 
smaller. 

By dividing the total stress acting on a certain area by the 
entire area, we obtain the mean intensity of the stress for the 
entire area. 

§ 179. Graphical Representation of Stress A conven- 
ient mode of representing stress 
graphically is the following: — 

Let AB (Fig. 163) be the plane 
surface upon which the stress acts ; 
let the axes OX and O Y be taken 
in this plane, the axis OZ being at 
right angles to the plane. 

Conceive a portion of a cylinder 
whose elements are all parallel to 
OZ, bounded at one end by the 
given plane surface, and at the FlG - 16 3- 

other by a surface whose ordinate at any point contains as 
many units of length as there are units of force in the intensity 
of the stress at that point of the given plane surface where the 
ordinate cuts it. 

The volume of such a figure will evidently be 

V = ffzdxdy = ffpdxdy, 




where z = / = intensity of the stress at the 



252 



APPLIED MECHANICS. 




Hence the volume of the cylindrical figure will contain as 
many units of volume as the total stress contains units of 
force ; or, in other words, the total stress will be correctly repre- 
sented by the volume of the body. 

If the stress on the plane 
figure is partly tension and 
partly compression, the sur- 
face whose ordinates repre- 
sent the intensity of the 
stress will lie partly on one 
side of the given plane sur- 
face and partly on the other ; 
this surface and the plane 
surface on which the stress 
acts, cutting each other in 
some line, straight or curved, 
as shown in Fig. 164. In that 
case, the magnitude of the resultant stress P- F= ffzdxdy 
will be equal to the difference of the wedge-shaped volumes 
shown in the figure. 

It will be observed that the above method of representing 
stress graphically represents, i°, the intensity at each point of 
the surface to which it is applied ; and, 2°, the total amount 
of the stress on the surface. It does not, however, represent 
its direction, except in the case when the stress is normal to 
the surface on which it acts. 

In this latter case, however, this is a complete representa- 
tion of the stress. 

The two most common uses of stress are, i°, uniform stress, 
and, 2°, uniformly varying stress. These two cases are repre- 
sented respectively in Figs. 165 and 166; the direction also 
being correctly represented when, as is most frequently the 
case, the stress is normal to the surface of action. In Fi<r. 
165, AD is supposed to be the surface on which the stress 



Fig. 164. 



GRAPHICAL REPRESENTATION OF STRESS. 



253 



acts ; the stress is supposed to be uniform, and normal to the 

surface on which it acts ; the bound- 
ing surface in this case becomes a 

plane parallel to AB ; the intensity 

of the stress at any point, as P, will 

be represented by PQ; while the 

whole cylinder will contain as many 

units of volume as there are units of 

force in the whole stress. 

Fig. 166 represents a uniformly 

varying stress. Here, again, AB is 

the surface of action, and the stress 

is supposed to vary at a uniform rate 

from the axis OY. The upper bounding surface of the cylin- 
drical figure which represents the stress 
becomes a plane inclined to the XOY 
plane, and containing the axis OY. 

In this case, if a represent the in- 
tensity of the stress at a unit's distance 
from OY, the stress at a distance x from 
OY will be/ = ax, and the total amount 
of the stress will be 




Fig. 165. 





^) 


<2^\ 


^3 




/ 


P ~~ N 




B 


\ 


/ 


A / 



Fig. 166. 



P = f fpdxdy = affxdxdy. 



When a stress is oblique to the surface of action, it may be 
represented correctly in all particulars, except in direction, in 
the aboye-stated way. 

§ 180. Centre of Stress. — The centre of stress, or the 
point of the surface at which the resultant of the stress acts, 
often becomes a matter of practical importance. If, for con- 
venience, we employ a system of rectangular co-ordinate axes, 
of which the axes OX and OY are taken in the plane of the 
surface on which the stress acts, and if we let / = cf>(x, y) be 



254 APPLIED MECHANICS. 

the intensity of the stress at the point (x, y), we shall have, 
for the co-ordinates of the centre of stress, 

_ Jfxpdxdy _ Jfypdxdy 

Xl ~ fjpdxdy ' yi ~ ffpdxdy ' 

(see § 42), where the denominator, or ffpdxdy, represents the 
total amount of the stress. 

When the stress is positive and negative at different parts 
of the surface, as in Fig. 135, the case may arise when the posi- 
tive and negative parts balance each other, and hence the 
stress on the surface constitutes a statical couple. In that case 

ffpdxdy = o. 

§ 181. Uniform Stress. — In the case of uniform stress, we 
have — 

i°. The intensity of the stress is constant, or / = a con- 
stant. 

2°. The volume which represents it graphically becomes a 
cylinder with parallel and equal bases, as in Fig. 165. 

3 . The centre of stress is at the centre of gravity of the 
surface of action ; for the formulae become, when / is constant, 

_ pffxdxdy _ ffxdx dy _ 
pffdxdy " ffdxdy 

= pffydxdy _ ffydxdy = 
yi ' pffdxdy " ffdxdy y » 

where x of y of are the co-ordinates of the centre of gravity of the 
surface. 

Examples of uniform stress have already been given in the 
cases of direct tension, direct compression, and, in the case of 
riveted joints, for the shearing-force on the rivet. 



UNIFORMLY VARYING STRESS. 255 

§182. Uniformly Varying Stress. — Uniformly varying 
stress has already been defined as a stress whose intensity varies 
uniformly from a given line in its own plane ; and this line will 
be called the Neutral Axis. Thus, if the plane be taken as the 
XOY plane (Fig. 166), and the given line be taken as OY, we 
shall have, if a denotes the intensity of the stress at a unit's 
distance from OY, and x the distance of any special point from 
O Y y that the intensity of the stress at the point will be 

p == ax. 

The total amount of the stress will be 

P = affxdxdy. 

The total moment of the stress about OY will be found by 
multiplying each elementary stress by its leverage. This lever- 
age is, in the case of normal stress, x ; hence in that case the 
moment of any single elementary force will be 

(ax Ax Ay) x = ax 2 Ax Ay, 

and the total moment of the stress will be 

M = affx 2 dxdy. 

In the case of oblique stress, this result has to be modified, 
as the leverage is no longer ;tr. Confining ourselves to stress 
normal to the plane of action, we have, for the co-ordinates of 
the centre of stress, 



yi = 
since 



where x OJ y Q , are the co-ordinates of the centre of gravity, and 
A is the area of the surface of action. 



ffpxdxdy affx 2 dxdy ffx 2 dxdy 
ffpdxdy P ffxdxdy 


ffx 2 dxdy 

x Q A 


ffpydxdy affxydxdy ffxydxdy 
ffpdxdy P ffxdxdy 


ffxydxdy 
XoA 


-P = affxdxdy == aXoA, 





256 



APrLIED MECHANICS. 



§ 183. Case of a Uniformly Varying Stress which 
amounts to a Statical Couple. — Whenever P = o, we have 



affxdxdy = o 



ffxdxdy — o 



x Q /i 



X n = O. 



In this case, therefore, we have — 

i°. There is no resultant stress, and hence the whole stress 
amounts to a statical couple. 

2°. Since x Q = o, the centre of gravity of the surface of 
action is on the axis OY, which is the neutral axis. 

Hence follows the proposition : — 

When a uniformly varying stress amounts to a statical coiLple, 
the neutral axis contains {passes through) the centre of gravity 
of the surface of action. 

In this case there is no single resultant of the stress ; but 
the moment of the couple will be, as has been already shown, 

M = affx 2 dxdy. 

§ 184. Example of Uniformly Varying Stress. — One of 
the most common examples of uniformly varying stress is that 
of the pressure of water upon the sides of the vessel contain- 



Thus, let Fig. 167 represent the vertical cross-section of a 
reservoir wall, the water pressing against the 
vertical face AB. It is a fact established by 
experiment, that the intensity of the pressure 
of any body of water at any point is propor- 
tional to the depth of the' point below the 
free upper level of the water, and normal to 
the surface pressed upon. Hence, if we sup- 
pose the free upper level of the water to be 
even with the top of the wall, the intensity 
of the pressure there will be zero ; and if we represent by CB 
the intensity of the pressure at the bottom, then, joining /2 and 





n 

B 


1 V 




\ 




| \ 




\ 




-r]-\ 


/e 


1 A 




1 \ 




1 1 \ 








1 1 \ 








1 ' \ 




1 i\ 



Fig. 167. 



STRESSES IN BEAMS UNDER TRANSVERSE LOAD. 2$7 



as 



C we shall have the intensity of the pressure at any point, 
D, represented by ED, where 

ED : CB = AD : AB. 



Here, then, we have a case of uniformly varying stress nor- 
mal to the surface on which it acts. 

§ 185. Fundamental Principles of the Common Theory 
of the Stresses in Beams under 
a Transverse Load. — Fig. 168 
shows a beam fixed at one end and 
loaded at the other, while Fig. 169 
shows a beam supported at the 
ends and loaded at the middle. 
Let, in each case, the plane of the 
paper contain a vertical longi- 
tudinal section of the beam. In 
o Fig. 168, 
fl it is evi- 
[J dent that 
the upper 



/ 

II 





Fig. 169. 



fibres are lengthened, while the lower 
ones are shortened, and vice versa in 
Fig. 169. In either case, there is, 
somewhere between the upper and 
lower fibres, a fibre which is neither 
elongated nor com- 
pressed. 

Let CN repre- 
sent that fibre, Fig. 

168, and CP, Fig. 

169. This line may 
be called trie neutral 
line of the longitu- 



dinal section ; and, if a section be made at any point at right 



258 APPLIED MECHANICS. 

angles to this line, the horizontal line which lies in the cross- 
section, and cuts the neutral lines of all the longitudinal sec- 
tions, or, in other words, the locus of the points where the 
neutral lines of the longitudinal sections cut the cross-section, 
is called the Neutral Axis of the cross-section. In the ordinary 
theory of the stresses in beams, a number of assumptions are 
made, which will now be enumerated. 

ASSUMPTIONS MADE IN THE COMMON THEORY OF BEAMS. 

Assumption No. i. — If, when a beam is not loaded, a 
plane cross-section be made, this cross-section will still be a 
plane after the load is put on, and bending takes place. From 
this assumption, we deduce, as a consequence, that, if a certain 
cross-section be assumed, the elongation or shortening per unit 
of length of any fibre at the point where it cuts this cross-sec- 
tion, is proportional to the distance of the fibre from the neutral 
axis of the cross-section. 

Proof. — Imagine two originally parallel cross-sections so 
near to each other that the curve in which that part of the 
neutral line between them bends may, without appreciable error, 
be accounted circular. Let ED and GH (Fig. 168 or Fig. 169) 
be the lines in which these cross-sections cut the plane of the 
paper, and let O be the point of intersection of the lines ED 
and GH. Let OF = r, EL = y, FK = /, LM = I + a/, in 
which a is the strain or elongation per unit of length of a fibre 
at a distance y from the neutral line, y being, a variable ; then, 
because FK and LM are concentric arcs subtending the same 
angle at the centre, we shall have the proportion 

r -h y I + a/ y 

or 1 + a = 1 H — 1 



r I r 

y 



or a 



G> 



ASSUMPTIONS IN THE COMMON THEORY OF BEAMS. 259 

but as y varies for different points in any given cross-section, 
while r remains the same for the same section, it follows, that, 
if a certain cross-section be assumed, the strain of any fibre at 
the point where it cats this cross-section is proportional directly 
to the distance of this fibre from the neutral axis of the cross- 
section. 

Assumption No. 2. — This assumption is that commonly 
known as Hooke s Law. It is as follows : " Ut tensio sic vis ; " 
i.e., The stress is proportional to the strain, or to the elonga- 
tion or compression per unit of length. As to the evidence in 
favor of this law, experiment shows, that, as long as the mate- 
rial is not strained beyond safe limits, this law holds. Hence, 
making these two assumptions, we shall have : At a given 
cross-section of a loaded beam, the direct stress on any fibre 
varies directly as the distance of the fibre from the neutral axis. 
Hence it is a uniformly varying stress, and we may repre- 
sent it graphically as follows : Let 
ABCD, Fig. 170, be the cross-sec- 
tion of a beam, and KL the neutral 
axis. Assume this for axis OY t and 
draw the other two axes, as in the 
figure. If, now, EA be drawn to 
represent the intensity of the direct 
(normal) stress at A y then will the 
pair of wedges AEFBKL and 
DCHGKL represent the stress graphically, since it is uni- 
formly varying. 

POSITION OF NEUTRAL AXIS. 

Assumption No. 3. — It will next be shown, that, on the 
two assumptions made above, and from the further assumption 
that the only resistances opposed to the bending of the beam 




260 



A PPL I ED ME CHA NICS. 



are the direct tensions and compressions of the fibres, it fol- 
lows that the neutral axis must pass through the centre of 
gravity of the cross-section. 




Fig. 



Fig. 172. 



Since the curvatures in Figs. 168 and 169 are exaggerated 
in order to render them visible, Figs. 171 and 172 have been 
drawn. If, now, we assume a section DE, such that AD = x 
(Fig. 171) and NE = x (Fig. 172), and consider all the forces 
acting on that part of the beam which lies to the right of DE 
(i.e., both the external forces and the stresses which the other 
parts of the beam exert on this part), we must find them in 
equilibrium. The external forces are, in Fig. 172, — 

i°. The loads acting between B and E ; in this case there 
are none. 

2°. The supporting force at B ; in this case it is equal to 

— , and acts vertically upwards. 

In Fig. 143 they are, — 

The loads between D and N ; in this case there is only the 
one, W at N. 

The internal forces are merely the stresses exerted by the 
other parts of the beam on this part : they are, — 

i°. The resistance to shearing at the section, which is a 
vertical stress. 

2°. The direct stresses, which are horizontal. 

Now, since the part of the beam to the right of DE is at 
rest, the forces acting on it must be in equilibrium ; and, since 



POSITION OF NEUTRAL AXIS. 26 1 

they are all parallel to the plane of the paper, we must have 
the three following conditions ; viz., — 

i°. The algebraic sum of the vertical forces must be zero. 

2°. The algebraic sum of the horizontal forces must be zero. 

3 . The algebraic sum of the moments of the forces about 
any axis perpendicular to the plane of the paper must be 
zero. 

But, on the above assumptions, the only horizontal forces 
are the direct stresses : hence the algebraic sum of these direct 
stresses must be zero ; or, in other words, the direct stresses 
must be equivalent to a statical couple. 

Now, it has already been shown, that, whenever a uniformly 
varying stress amounts to a statical couple, the neutral axis 
must pass through the centre of gravity of the surface acted 
upon. Hence in a loaded beam, if the three preceding assump- 
tions be made, it follows that the neutral axis of any cross- 
section must contain the centre of gravity of that section. 

By way of experimental proof of this conclusion, Barlow 
has shown by experiment, that, in a cast-iron beam of rectangu- 
lar section, the neutral axis does pass through the centre of 
gravity of the section. 

RESUME. 

The conclusions arrived at from the foregoing are as fol- 
lows : — 

i°. That at any section of a loaded beam, if a horizontal 
line be drawn through the centre of gravity of the section, 
then the fibres lying along this line will be subjected neither 
to tension nor to compression ; in other words, this line will be 
the neutral axis of the section. 

2°. The fibres on one side of this line will be subjected to 
tension, those on the other side being subjected to compres- 
sion ; the tension or compression of any one fibre being propor- 
tional to its distance from the neutral axis. 



262 APPLIED MECHANICS. 

§ 1 86. Shearing-Force and Bending-Moment. — In deter- 
mining the strength of a beam, or the proper dimensions of a 
beam to bear a certain load, when we assume the neutral axis 
to pass through the centre of gravity of the cross-section, we 
have imposed the second of the three last-mentioned conditions 
of equilibrium. The remaining two conditions may otherwise 
be stated as follows : — 

i°. The total force tending to cause that part of the beam 
that lies to one side of the section to slide by the other part, 
must be balanced by the resistance of the beam to shearing at 
the section. 

2°. The resultant moment of the external forces acting on 
that part of the beam that lies to one side of the section, about 
a horizontal axis in the plane of the section, must be balanced 
by the moment of the couple formed by the resisting stresses. 

The shearing-force at any section is the force with which the 
part of the beam on one side of the section te?ids to slide by the 
part on the other side. In a beam free at one end, it is equal to 
the sum of the loads between the section and the free end. In 
a beam supported at both ends, it is equal in magnitude to the 
difference between the supporting force at either end, and 
the sum of the loads between the section and that support. 

The bending-moment at any section is the resultant moment 
of the external forces acting on the part of the beam to one 
side of the section, these moments being taken about a hori- 
zontal axis in the section. 

In a beam free at one end, it is equal to the sum of the 
moments of the loads between the section and the free end, 
about a horizontal axis in the section. 

In a beam supported at both ends, it is the difference be- 
tween the moment of either supporting force, and the sum of 
the moments of the loads between the section and that sup- 
port ; all the moments being taken about a horizontal axis in 
the section. 



SHEARING-FORCE AND B ENDING-MOMENT. 263 

Hence the two conditions of equilibrium may be more 
briefly stated as follows : — 

i°. The shearing-force at the section must be balanced 
by the resistance opposed by the beam to shearing at the 
section. 

2°. The bending-moment at the section must be balanced 
by the moment of the couple formed by the resisting stresses. 

It is necessary, therefore, in determining the strength of a 
beam, to be able to determine the shearing-force and bending- 
moment at any, point, and also the greatest shearing-force and 
the greatest bending-moment, whatever be the loads. 

A table of these values for a number of ordinary cases will 
now be given ; but I should recommend that the table be merely 
considered as a set of examples, and that the rules already 
given for finding them be followed in each individual case. 

Let, in each case, the length of the beam be /, and the 
total load W. When the beam is fixed at one end and free at 
the other, let the origin be taken at the fixed end ; when it is 
supported at both ends, let it be taken directly over one support. 
Let x be the distance of any section from the origin. Then we 
shall have the results given in the following table : — 



264 



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MOMENTS OF INERTIA OF SECTIONS. 26$ 

In a beam fixed at one end and free at the other, the great- 
est shearing-force, and also the greatest bending-moment, are at 
the fixed end. In a beam supported at both ends, and loaded 
at the middle, or with a uniformly distributed load, the greatest 
shearing-force is at either support, the greatest bending-moment 
being at the middle. In the last case (i.e., that of a beam sup- 
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greatest. 

§ 187. Moments of Inertia of Sections. — In the usual 
methods of determining the strength of a beam or column, it 
is necessary to know, i°, the distance from the neutral axis of 
the section to the most strained fibres ; 2°, the moment of in- 
ertia of the section about the neutral axis. The manner of 
finding the moments of inertia has been explained in Chap. II. 

In the following table are given the areas of a large number 
of sections, and also their moments of inertia about the neutral 
axis, which is the axis YY in each case. These results should 
be deduced by the student. 



266 



APPLIED MECHANICS. 



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276 



APPLIED MECHANICS. 



§ 188. Cross-Sections of Phoenix Columns considered 
as made of Lines. — It is to be observed that the moments 
of inertia are the same for all axes passing through the centre. 
Thickness 5= t, radius of round ones = r, area of each flange 
= a y length of each flange — /. 



Figure. 



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Description. 



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REPRESENTATION OF B ENDING-MOMENTS. 



277 




§ 189. Graphical Representation of Bending-Moments. — 

The bending-moment at each point of a loaded beam may be 
represented graphically by lines laid off to scale, as will be 
shown by examples. 

I. Suppose we have the cantilever shown in Fig. 215, 
loaded at D with a load W : then 
will the bending-moment at any 
section, as at F f be obtained by 
multiplying W by FD ; that at AC 
being W X (AB). If, now, we lay 
off CE to scale to represent this, 
i.e., having as many units of length 
as there are units of moment in the product W X (AB), and 
join E with D, then will the ordinate FG of any point, as G, 
represent (to the same scale) the bending-moment at a section 
through F 

II. If we have a uniformly distributed load, we should have, 
for the line corresponding to CE in Fig. 215, a curve. This is 

shown in Fig. 216, where we have the 
uniformly distributed load EIGF. If 
we take the origin at D, as before, 
we have, for the bending-moment, at a 
distance x from the origin, as has been 



Fig. 215. 




Fig. 216. 



w 

shown, — (/ 
2/ 



x) 2 ; and by giving x dif- 



ferent values, and laying off the corresponding value of the 
bending-moment, we obtain the curve CA, any ordinate of 
which will represent the bending-moment at the corresponding 
point of the beam. 

When we have more than one load on a beam, we must draw 
the curve of bending-moments for each load separately, and 
then find the actual bending-moment at any point of the beam 



278 



APPLIED MECHANICS. 



by taking the sum of the ordinates (drawn from that point) of 
each of these separate curves or straight lines. If we then 
draw a new curve, whose ordinates are these sums, we shall have 
the actual curve of bending-moments for the beam as loaded. 
Some examples will now be given, which will explain them- 
selves. 



III. Fig. 2 




7 shows a cantilever with three concentrated 
loads. The line of bending-moments 
for the load at C is CE, that for the 
load at O is OF, and for the load at P 
is PG. They are combined above the 
beam by laying off AH = BE, HK = 
DF, and KL = DG, and thus obtaining 
the broken line LMNB, which is the 
line of bending-moments of the beam 
loaded with all three loads. 



Fig. 217. 



IV. Fig. 218 shows the case of a beam supported at both 
ends, and loaded at a single point 
D; ALB is the line of bending- 
moments when the weight of the 
beam is disregarded, so that xy = 



bending-moment at x. 




Fig. 218. 



V. Fig. 219 shows the case of a beam supported at the ends, 
and loaded with three concentrated 
loads at the points B, C, and D re- 
spectively ; the lines of bending-mo- 
ments for each individual load being 
respectively AFE, AGE, and AHE, 
Fi<;. 219- and the actual line of bending-mo- 

ments being AKLME. 




REPRESENTATION OF B ENDING-MOMENTS. 



279 



VI. Fig. 220 shows the case of a beam supported at the 
ends, and loaded with a uniformly dis- 
tributed load ; the line of bending- 
moments being a curve, ACDB, as 
shown in the figure. 





VII. In Fig. 221 we have the case of a beam, over a part of 
which, viz., EF y there is a distributed load ; the rest of the 

beam being unloaded. The line of 
bending-moments is curvilinear be- 
tween E and F, and straight outside 
of these limits. It is AG SHB ; and, 
when the curve is plotted, we can 
find the greatest bending-moment 
graphically by finding its greatest ordinate. We can also 
determine it analytically by first determining the bending- 
moment at a distance x from the origin, and on the side 
towards the resultant of the load, and then differentiating. 
This process is shown in the following: — 
Let A (Fig. 222) be the point where 
the resultant of the load acts, and O the 
middle of the beam, and let w be the 
load per unit of length ; let OA — a, and 
AB = AC = b, so that the whole load == 2wb : therefore sup- 

P ^ , a "4" c wb(a -f- c) 
porting force at D = 2wb = . 

If we take a section at a distance x from O to the right, we 
shall have, for the bending-moment at that section, 





1 1 1 




E 


1 1 


D 


1 1 1 


1 . 1 1 



C OA 
Fig. 222. 



wb(a -\- c) , x W 

e •■■■' (* -x) -~{a + i>- 

Differentiate, and we have 

-wb(a + c) 

f- w\a +b — x) = o 



x) 



a maximum. 



a(c — b) 



28o 



APPLIED MECHANICS. 



hence the greatest bending-moment will be 



wb(a -\- c) 



ab\< 



±^{c- a{C ~ *) -^(a + t-a + ^y 



= — (a -f c) \c 2 — ac + ao) 



wb z 

2C 2 



{a 2 -f 2dc -f C 2 ) 



wb 



= —{a 2 b — 2a 2 c -f- 2<r 3 — ^ 2 ), 

2<T 2 



VIII. In Figs. 223 and 224 we have the case of a beam 
!> supported at 

the ends, and 

loaded with a 

uniformly dis- 
8 tributed load, 

and also with 

a concen- 
trated load. 

In the first 

figure, the greatest bending-moment 





Fig. 223. 



is at D, and in the second at C. 



IX. In Fig. 225 we have a beam supported at A and B, and 
loaded at C and D with equal 
weights ; the lengths of AC and 
BD being equal. We have, con- 
sequently, between A and B, a 
uniform bending-moment ; while 
on the left of A and on the right 
of B we have a varying bending-moment. 
moments is, in this case, CabD. 

We may, in a similar way, derive curves of bending-moment 
for all cases of loading and supporting beams. 




Fig. 225. 

The line of bending- 



STRESSES AT DIFFERENT PARTS OF A BEAM. 28 I 

§ 190. Mode of Procedure for Ascertaining the Stresses 
at Different Parts of a Beam when the Loads and the Di- 
mensions are given. — When the dimensions of a beam, the 
load and its distribution, and the manner of supporting are 
given, and it is desired to find the actual intensity of the stress 
on any particular fibre at any given cross-section, we must pro- 
ceed as follows : — 

i°. Find the actual bending-moment (M) at that cross-sec- 
tion. 

2°. Find the moment of inertia (/) of the section about its 
neutral axis. 

3 . Observe, that, from what has already been shown, the 
moment of the couple formed by the tensions and compressions 
is al, where a = intensity of stress of a fibre whose distance 
from the neutral axis is unity, and that this moment must equal 
the bending-moment at the section in order to secure equilib- 
rium. Hence we must have 

af= M. 

Moreover, if / denote the (unknown) intensity of the stress 
of the fibre where the stress is desired, and if y denote the 
distance of this fibre from the neutral axis, we shall have 

•4 

from which equation we can determine/. 

EXAMPLES. 

i. Given a beam 18 feet span, supported at both ends, and loaded 
uniformly (its own weight included) with 1000 lbs, per foot of length. 
The cross-section is a T, where area of flange = 3 square inches, 
area of web = 4 square inches; height =10 inches. Find (a) the 



282 APPLIED MECHANICS. 

bending at 3 feet from one end ; {b) the greatest bending-moment ; 

(c) the greatest intensity of the tension at each of the above sections ; 

(d) the greatest intensity of the compression at each of these sections. 
2. Given an I-beam with equal flanges, area of each flange = 3 

square inches, area of web = 3 square inches, height = 10 inches; the 
beam is 12 feet long, supported at the ends, and loaded uniformly (its 
own weight included) with a load of 2000 lbs. per foot of length. Find 
(a) the bending-moment at a section one foot from the end ; (l>) the 
greatest bending-moment ; (V) the greatest intensity of the stress at 
each of the above bross-sections. 



§ 191. Mode of Procedure for Ascertaining the Dimen- 
sions of a Beam to bear a Certain Load, or the Load that 
a Beam of Given Dimensions and Material is Capable of 
Bearing. — If we wish to determine the proper dimensions 
of the beam when the load and its distribution, as well as the 
manner of supporting, are given, so that it shall nowhere be 
strained beyond safe limits, or if we wish to determine the 
greatest load consistent with safety when the other quantities 
are given, we must impose the condition that the greatest 
intensity of the tension to which any fibre is subjected shall 
not exceed the safe working-strength for tension of the mate- 
rial of which the beam is made, and the greatest intensity of 
the compression to which any fibre is subjected shall not exceed 
the safe working-strength of the material for compression. 

Thus, we must in this case first determine where is the 
section of greatest bending-moment (this determination some- 
times involves the use of the Differential Calculus). 

Next we must determine the magnitude of the greatest 
bending-moment, absolutely if the load and length of the beam 
are given (if not, in terms of these quantities), and then equate 
this to the moment of the resisting couple. 

Thus, if M Q is the greatest bending-moment, I the moment 
of inertia of that section where this greatest bending-moment 



WORKING-STRENGTH. 283 

acts, and \i f t =. working-strength per square inch for tension, 
f c z=. working-strength per square inch for compression, y t = 
distance of most stretched fibre from the neutral axis, and y c 
= distance of most compressed fibre from the neutral axis, 

then will — be the greatest tension per square inch, at a unit's 

yt 

distance from the neutral axis, consistent with safety against 
tearing, and — the greatest compression per square inch, at a 

yc 

unit's distance from the neutral axis, consistent with safety 
against crushing. 

Of course the least of these must not be exceeded in the 
actual beam. Hence we must put 

M» = f -I, 
where - is taken as the lesser of the two quantities — and — . 

y yt y c 



MODULUS OF RUPTURE. 

The modulus of rupture is the greatest tension or com- 
pression per square inch to which the most strained fibre of 
the beam is subjected when the beam is just on the point 
of breaking. 

WORKING-STRENGTH. 

The working-strength per square inch of a material for 
transverse strength is the greatest stress per square inch to 
which it is safe to subject the most strained fibre of the beam. 
It is usually obtained by dividing the modulus of rupture by 
some factor of safety, as 3 or 4. 



284 APPLIED MECHANICS. 



§ 192. EXAMPLES. 

i. Given, as the modulus of rupture of a spruce beam, 4000 lbs. per 
square inch : find its breaking-strength, assuming it to be 4 inches wide 
and 12 inches deep, the span being 18 feet, the load being uniformly 
distributed over its entire length. 

2. Suppose such a beam to break with a load at the middle of 5000 
lbs. : find its modulus of rupture. 

3. Given a T-beam fixed at one end, and loaded uniformly. The 
area of the flange is 3 square inches, that of the web also 3 square 
inches, height = 10 inches. The beam is 4 feet long. Find the greatest 
load it will bear with safety, the working-strength per square inch of the 
material being, for tension, 10000 lbs., and for compression 8000 lbs. 

4. Given an I-beam, area of each flange being 3 square inches, and 
area of web 3 square inches, height = 12 inches, span 8 feet, supported 
at the ends, and loaded uniformly : what load will it bear with safety, the 
working-strength of the iron for tension, and also for compression, being 
12000 lbs. per square inch. 

5. Given a beam (Fig. 226) supported at both ends, and loaded, 
i°, with w pounds per unit of length uniformly, and 2°, with a single 
load W at a distance a from the left-hand support : find the position 
of the section of greatest bending-moment, and the value of the greatest 
bending- moment. 

Solution. 
o a b 1. Left-hand supporting- force 



^ _ w/ W(I - a) 

7 2 7 ' 

Right-hand supporting-force 

Fig. 226. wl Wa 

= 7 + ~T' 

2. Assume a section at a distance x from the left-hand support 
(this support being the origin), and the bending-moment at that 
section is, — 



when x < a, 



and when x > a, 



WORKING-STRENGTH. 285 



wl W{1 — a) ) wx 2 

2 / ) 2 



To find the value of x for the section of greatest bending-moment, 
differentiate each, and put the first differential co-efficient = zero. 
We shall thus have, in the first case, 

wl W(l -a) I W(J - a) 

1 -. wx = o, or x — - -\ -j ; 

2 / 2 wl 

and in the second case, 

wl W(l -a) I W(l -a) W 

j — wx — W — o, or x — — I -j . 

2 . / 2 wl w 

Now, whenever the first is < a, or the second is > a, we shall have 
in that one the value of x corresponding to the section of greatest 
bending-moment. But if the first is > a, and the second < a, then the 
greatest bending-moment is at the concentrated load. 

These conclusions will be evident on drawing a diagram 
representing the bending-moments graphically, as in Figs. 223 
and 224 ; and the greatest bending-moment may then be found 
by substituting, in the corresponding expression for the bend- 
ing-moment, the deduced value of x. 

6. Given an I-beam 10 feet long, supported at both ends, and 
loaded, at a distance 2 feet to the left of the middle, with 20000 pounds. 
Find the bending-moment at the middle, the greatest bending-moment, 
also the greatest intensity of the tension, and that of the compression at 
each of these sections. 

Given Area of upper flange = 8 sq. in. 

Area of lower flange = 5 sq. in. 

Area of web = 7 sq. in. 
Total depth = 14 in. 



286 APPLIED MECHANICS. 

7. Given a beam (Fig. 227) 18 feet long, loaded at A with 1000 

o a b c lbs., and at B with 2000 lbs. ; the beam 

^ weighing 200 lbs. (OA = 3 feet, OB 
= 10 feet). Find the section of great- 
est bending-moment, and the bending- 
Ofl moment at that section. 

Fig. 227. 

§ 193. Beams of Uniform Strength. — Abeam of uniform 
strength (technically so called) is one in which the dimensions 
of the cross-section are varied in such a manner, that, at each 
cross-section, the greatest intensity of the tension shall be 
the same, and so also the greatest intensity of the com- 
pression. 

Such beams are very rarely used ; and, as the cross-section 
varies at different points, it would be decidedly bad engineering 
to make them of wood, for it would be necessary to cut the 
wood across the grain, and this would develop a tendency to 
split. 

In making them of iron, also, the saving of iron would gen- 
erally be more than offset by the extra cost of rolling such a 
beam. Nevertheless, we will discuss the form of such beams in 
the case when the section is rectangular. 

In all cases we have the general equation 

y 

applying at each cross-section, where M =2 bending-moment 
(section at distance x from origin), / = moment of inertia of 
same section, y = distance from neutral axis to most strained 
fibre, and p = intensity of stress on most strained fibre ; the 
condition for this case being that / is a constant for all values 
of x (i.e., for all positions of the section), while M t I, and y 
are functions of x. 



BEAMS OF UNIFORM STRENGTH. 287 

As we are limiting ourselves to rectangular sections, if we 
let b = breadth and h = depth of rectangle (one or both vary- 
ing with x) t we shall have 



as the condition for such a beam, with/ a Constant for all values 
of x, when the same load remains on the beam. 

We must, therefore, have bh 2 proportional to M. Hence, 
assuming the origin as before, 

i°. Fixed at one end, load at the other, bh 2 = (—) W{1 — x). 

2 . Fixed at one end, uniformly loaded, bh 2 = ( ) (/— x) 2 . 

\p 2/' 

3 . Supported at ends, loaded at I 2 \p 2 / 

middle - Ifor^M'^-^K/-*). 

V 2 \p 2 / 

4°. Supported at ends, uniformly loaded, bh 2 —\ ){lx — x 2 ). 

\p 2// 

Now, this variation of section may be accomplished in one 
of two ways: 1st, by making h constant, and letting b vary; 
and 2d, by making b constant, and letting h vary. Thus, in 
the first case above mentioned,if h is constant, we have, for the 
plan of the beam, 

and if one side be taken parallel to the axis of the beam, this 
will be the equation of the other si<sle ; and, as this is the equa- 
tion of a straight line, the plan will be a triangle. 



288 APPLIED MECHANICS. 

If, on the other hand, b be constant, and h vary, we shall 
have, for the vertical longitudinal section of the beam, 



* = ('-£)«->; 



pb 

and, if one side be taken as a straight line in the direction of 
the axis, the other will be a parabola. 

A similar reasoning will give the plan or elevation respect- 
ively in each case ; and these can be readily plotted from their 
equations. 

CROSS-SECTION OF EQUAL STRENGTH. 

A cross-section of equal strength (technically so called) is 

one so proportioned that the greatest intensity of the tension 

shall bear the same ratio to the breaking tensile strength of the 

material as the greatest intensity of the compression bears to 

the breaking compressive strength of the material. This is 

accomplished, as will be shown directly, by so arranging the 

form and dimensions of the section that the distance of the 

neutral axis from the most stretched fibre shall bear to its 

distance from the most compressed fibre the same ratio that 

the tensile bears to the compressive strength of the material. 

Let fi — breaking-strength per square inch for compression, 

f t = breaking-strength per square inch for tension, 

y c = distance of neutral axis from most compressed 

fibre, 

y t = distance of neutral axis from most stretched fibre. 

If p c — actual greatest intensity of compression, and p t = 

actual greatest intensity of tension, then, for a cross-section 

of equal strength, we must have, according to the definition, 

<^= *-\ but we have — = — = intensity of stress at a unit's 
Pt ft y c yt 



CROSS-SECTION OF EQUAL STRENGTH. 289 

distance from the neutral axis. Hence, combining these two, 
we obtain 

y ± =&. 

yt ft 

EXAMPLE. 

Suppose we have/ c = 80000 lbs. per square inch, andy* = 20000 
lbs. per square inch. : find the proper proportion between the flange A t 
and the web A 2 of a T-section whose depth is h. 

§ 194. Deflection of Beams. — We have already seen (§ 185), 
that, in the case of a beam which is bent by a transverse load, 
we have 

a = IOO, 

where (having assumed a certain cross-section whose distance 
from the origin is x) a = the strain of a fibre whose distance 
from the neutral axis is f, and r — radius of curvature of 
the neutral lamina at the section in question. Hence follows the 
equation 

I _ a 

- r -y 

but from the definition of E f the modulus of elasticity, we shall 
have 

P 
a = E> 

where / = intensity of the stress at a distance y from the 
neutral axis. 

Hence it follows, assuming Hooke's law, that 

I = 1- = Li 
r Ey E y 

We have already seen, that, disregarding signs, M = - / 



29O APPLIED MECHANICS. 

(making, of course, the two assumptions already spoken of 
when this formula was deduced), where M = bending-moment 
at, and / = moment of inertia of, the section in question ; i.e., 
of that section whose distance from the origin is x. This gives 

<- = , if, denoting tension by the -f- sign, and taking y 

y ! 

positive upwards, we call M positive when it tends to cause 
tension on the lower, and compression on the upper, side; these 
being the conventions in regard to signs which we shall adopt 
in future. Hence, by substitution, we have 

1 = ± = _ M _ . (i) 

r Ey EI 

Now, if we assume the axis of x coincident with the neutral 
line of the central longitudinal section of the beam, and the 
axis of v at right angles to this, and v positive upwards, no 
matter where the origin is taken, we shall always have, as is 
shown in the Differential Calculus, 

_ d*v_ 
1 ~dx 2 



- ('-m? 



Hence equation (1) becomes 



d 2 v 

~dx z M 



(■♦(*)■) 



^Y\t ££ 



(2) 



M and / being functions of x : and, when we can integrate 
this equation, we can obtain v in terms of x, thr.s having the 
equation of the elastic curve of the neutral line ; and, by com- 
puting the value of v corresponding to any assumed value of x> 
we can obtain the deflection at that point of the beam. 



FORMULA FOR SLOPE AND DEFLECTION. 29 1 

The above equation (2) is, as a rule, too complicated to be 
integrated, except by approximation ; and the approximation 
usually made is the following : — 

Since in a beam not too heavily loaded, the slope, and con- 
sequently the tangent of the slope (or angle the neutral line 
makes with the horizontal at any point), is necessarily small, it 

follows that — is very small, and hence ( — ) is also very small, 
dx \dx/ 

and 1 -f- f — ) is nearly equal to unity. Making this substitu- 
tion, we obtain, in place of equation (2), 

dx* ~ El' Kl) 

and this is the .equation with which we always start in com- 
puting the slope and deflection of a loaded beam, or in finding 
the equation of the elastic line. 

By one integration (suitably determining the arbitrary con- 
stant) we obtain the slope whose tangent is — , and by a second 

dx 

integration we obtain the deflection v at a distance x from the 
origin ; and thus, by substituting any desired value for x, we 
can obtain the deflection at any point. 

§195. Ordinary Formulae for Slope and Deflection. — 
We may therefore write, if i is the circular measure of the 

slope at a distance x from the origin, since i = tan i = -~ 

dx 
nearly, 

d*v = M 

dx 2 Ef 



dv CM , 



EI 



\^dx\ 



II: 



292 APPLIED MECHANICS. 

In these equations, of course, E is taken as a constant, M 
must always be expressed in terms of x, and so also must / 
whenever the section varies at different points. When, how- 
ever, the section is uniform, / is constant, and the formulae 
reduce to 

! ' = i//*- v = hJS MdxK 

§ 196. Special Cases. — i°. Let us take a cantilever loaded 
with a single load at the free end. Assume the origin, as 
before, at the fixed end, and let the beam be one of uniform 
section. We then have M = — W(l — x), 

.*. 1 — I (/— x)dx = 1 Ix ) -+- c. 

EI J V ; EI\ 2 / T 

To determine c, observe that when x = o, i = o ; 

•'• { - -fM x - f) (i) 

is the slope at a distance x from the origin. 
The deflection at the same point will be 

•-/-- -£/(*-!>= -£(? -?)+« 

but when ^ = 0,^ = .'. c — o .'. the deflection at 
a distance x from the origin will be 



W/lx* x\ f \ 



The equations (1) and (2) give us the means of finding the 
slope and deflection at any point of the beam. 

To find the greatest slope and deflection, we have that both 
expressions are greatest when x = I. Hence, if i Q and v Q rep- 
resent the greatest slope and deflection respectively, 

2EI lEI 



SPECIAL CASES. 293 



2°. Next take the case of a beam supported at both ends 
and loaded uniformly, the load per unit of length being w. 
Assume the origin at the left-hand end ; then 

wl wx 2 w 

M = — x — = —Ux — x 2 ) and W — wl 

2 2 2 v 

w 



1 ~ 2EI 



/w fix 2 x*\ 



I 

To determine c, we have that when x = -, then i = o : 

2 

w // 3 / 3 \ wl 1 

•"• o = — Fr J- )+c .'. e= - 



2EI V 8 24/ 24EI 

w fix 2 x 3 \ wl 3 w 

/w C 

idx — — — I (6lx 2 — 4x* — l*)dx 



24EI 



(2/^3 - x* - A*) + c. 



But when x = o, v = o ; 



.". ^ = o 

•*• v = ' 2 ^eA 2/xz -**- fi *y- <*) 

Tor the greatest slope, we have x = o, or x = I; 

wl 3 Wl 2 
""" lo= 2~JE/ = ^Ef' 

I 
For the greatest deflection, # = - ; 

w 5/+ 5a// 4 5 #73 

^° ~ 24^/16" = 384^/ = 384^/' 



294 APPLIED MECHANICS. 

3°. Take the case of a beam supported at both ends, and 
loaded at the middle with a load W. 

Assume, as before, the origin at the left-hand support. 
Then we shall have 

W I W I 

M = —x, x < -, and M = — (/ — x) when x > — 
2 2 2 v y 2 

Therefore, for the slope up to the middle, we have 

W r W x 2 

i = —^J xdx = -tetj— + c 



2EIJ 2EI 2 



/ 

When x = -, then * = o ; 

2 



and 



7 

4EI 



w ( /2 \ 



W Ct /2 \ ^ /* 3 /2 *\ 

But when ;tr = o, z; = o ; 



<: = o 



W (x* Px\ 

V = —ptA )• (2) 

*EI V3 4 ' 



The slope is greatest when x = o ; 

#7 a 



16^/ 



/ 
The deflection is greatest when x = - ; 

0?3 



4&J5/ 

In this case the symmetry of the beam and load makes it 



SPECIAL CASES. 



295 



unnecessary to examine the part where x> -\ but, if this were 
to be done, we should have, for that part, 



W 

Tel 



I (/ - x)dx and v = j^-A J (' - *)**. 



4 . Following will be found a table of some deflections, 
which may be regarded as examples simply. 



Uniform Cross-Section. 


Greatest Slope. 


Greatest 
Deflection. 


Fixed at one end, loaded at the other 
Fixed at one end, loaded uniformly . . 
Supported at ends, load at middle . . . 
Supported at ends, uniformly loaded . . 


1 Wl 2 
2 ~EI 

1 wi 2 

&EI 

1 m* 

16 EI 

2 Wl 2 
2 AEI 


1 wis 

3 EI 

1 wis 

8 EI 
, Wis 

&EI 

5 Wis 
3*4 EI 



Uniform Strength and Uniform Depth. 



Fixed at one end, load at the other . . 
Fixed at one end, uniformly loaded . . 
Supported at both ends, load at middle . 
Supported at both ends, uniformly loaded, 



Wl 2 


1 Wis 


EI 


2 EI 


1 W/ 2 


1 Wis 


2 ~EI 


A EI 


1 Wl 2 


1 Wis 


8 EI 


T*~EI 


1 Wl 2 


1 Wis 


16 EI 


saeT 



296 APPLIED MECHANICS. 

§197. Deflection with Uniform Bending-Moment If 

the bending-moment is uniform, then M is constant ; and, if / 
is also constant, we have 



M r Mx 



I 
but when x = -, then 1 == o; 
2 

Ml 



2EI 

dv 
dx 



.Ml l\_dv 
'* m % ~ EI\ X ~~2}~~dx 

M(x 2 lx\ 



the constant disappearing because v = o when j = o. 

Hence, for a beam where the bending-moment is uniform, 
we have 



1 = 



J// A M/x> /x\ 

El\ X ~2} v = £l\l-lp 



and for greatest slope and deflection, we have 



-Ml Mil* 



l 2 \ _ 3 Ml 2 



§ 198. Resilience of a Beam. — The resilieiice of a beam 

is the mechanical work performed in deflecting it to the amount 

it would deflect under its greatest allowable gradually applied 

load. In the case of a concentrated load, if W is the greatest 

allowable gradually applied load, and v z the corresponding 

deflection at the point of application of the load, then will the 

.' W 
mean value of the load that produces this deflection be — » 

W 
and the resilience of the beam will be ~v x * 



SLOPE AND DEFLECTION OF A BEAM. 297 



§ 199. Slope and Deflection of a Beam with a Con 
centrated Load not at the < 

Middle Take, as the next °. __a b 

case, a beam (Fig. 228). Let < a 

the load at A be W, and dis- 
tance OA = a, and let a > -. 

2 Fig. 228. 



x < a M = — ^ -t at, 



# < a M = (/ — *), 



When x = o, * = 4 = undetermined slope at (9; 



and 



' = '°' ■■• * = a a?/ * + '°> (I) 



- g) J^ + 4 J^ = ^(/ -/) .3 + ^ + ,. 



2IEI J V 6ZE7 

When x = o, z/ = o ; 



" = ~kr * 3 + ** (2) 



IEL 



!/<'-«>* =£(*-?)+<■ 



To determine <:, observe that when ;r = a, this value of z 
and that deduced from (1) must be identical. 

Waf. a 2 \ W{1 - a) a? , . Wa* , . 

IEI\ 2 / 2IEI 2EI 



298 APPLIED MECHANICS. 



IEJ\ 2 J 



x 2 \ Wa 2 , . 



or 



i = -^§rf 2lx - & ~ la ) + 4, (3) 

and 

v = / (2/x — x 2 — la)dx + 4 I ak 

= —^-Alx 2 — lax) + 4* + c. 

2 IE I 3 

To determine c y observe that when x = a, this value of v 
and (2) must be identical ; 

Wa / a\ , . W(l-a) • t .- . 

6/£Y V ' 6/£7 6^7 

•*• » = 7^?(3^ 2 - * 3 - 3&* + ^ 2 ) + 'ax. (4) 
To determine i of we have that when x .= I, v = o; 

•"• ° = 77F"/ 2/3 ~ 3^ /2 + ^ 2 ) + V 

. = _Wa_( /2 _ 2/3 _ / - = J^( 3 ^/ _ 2 /> _ a >). 
6l 2 EJ X6 ' 61EI K * 

Substituting this value of i in the equations (1), (2), (3), and 
(4), we obtain for 

; ■ " W{1 - a) 2 Wa J2 2N 

(1) 1 = — -x 2 4- (xal — 2I 2 — # 2 ), 

V ; 2IEI 61E/ Kd J 

W(l - a) , «^z , . /2 2N 



SLOPE AND DEFLECTION OF A BEAM. 299 



(3) / - J^L( 2 /x - x 2 - a/) + -^-(3*/ - 2/2 - a 2 ), 

(4) v = &&* ~ xz ~ zlax + /a2) + M/ (sa/ ~ 2/2 ~ a2)x - 

To find the greatest deflection, differentiate (2), and place 
the first differential co-efficient equal to zero : or, which is the 
same thing, place i == O in (1), and find the value of x ; then 
substitute this value in (2), and we shall have the greatest 
deflection. 

We thus obtain 

(/ _ a)x > = ^L {za i _ 2/2 L a 2 ) ,. x 2 = *t**J=J£l±*\ 
3 3\ / - a I 

or 



x 2 = -(2/ — a) ,\ x = 



\2a/ — a" 



3 V^ 

and the greatest deflection becomes 



V Q = 



_ Wa{l - a) (2/ - a) Vial - a 2 
g/JSI ^ 



§ 200. EXAMPLES. 

1. In example 1, p. 284, find the greatest deflection of the beam 
when it is loaded with J of its breaking-load, assuming E = 1200000. 

2. In the same case, find what load will cause it to deflect ^Jq of its 
span. 

3. What will be the stress at the most strained fibre when this occurs. 

4. In example 3, p. 284, find the load the beam will bear without 
deflecting more than ^^ of its span, assuming E — 24000000. 

5. Find the stress at the most strained fibre when this occurs. 

6. In example 6, p. 285, find the greatest deflection under a load 
\ the breaking-load. 



300 



APPLIED MECHANICS. 



§ 201. Deflection and Slope under Working-Load. — If 
we take the four cases of deflection given in the- first part of 
the table on p. 295, and calling/" the modulus of rupture of the 
material, and y the distance of the most strained fibre from 
the neutral axis, and if we make the applied load the working- 
load, we shall have respectively — 



Wl = 


fl 

y 


Wl 

2 


fl 

' y 


Wl 


fl 


4 


y 


Wl 


fl 

y 



w=. 



w= 



w= 



w= 



fl 

2/1 

iy' 
iy' 
iy' 



And the values of slope and deflection will become respectively, 





Slope. 


Deflection. 




Slope. 


Deflection. 




/ 


/ 2 




/ 


fi 2 


1°. 


Vbj 


^Ey 


3"- 


Vjfy 


~^Ty 




I 


/ 2 




I 


I 2 


2 U . 


if 


v# 


4°- 


\f~Ey 


5 / 

** J Ey 



From these values, and those given on p. 295, we derive the 
following two propositions : — 

i°. If we have a series of beams differing only in length, 
and we apply the same load in the same manner to each, their 
greatest slopes will vary as the squares of their lengths, and 
their Greatest deflections as the cubes of their lengths. 



SLOPE AND DEFLECTION OF RECTANGULAR BEAMS. 3OI 

2°. If, however, we load the same beams, not with the same 
load, but each one with its working-load, as determined by 
allowing a given greatest fibre stress, then will their greatest 
slopes vary as the lengths, and their greatest deflections as the 
squares of their lengths. 

§ 202. Slope and Deflection of Rectangular Beams. — 

M 3 , h 

If the beams are rectangular, so that I = — and y = -, the 

values of slope and deflection above referred to become further 
simplified, and we have the following tables : — 





Given Load W. 


Working-Load. 
Greatest Fibre Stress =/. 




Slope. 


Deflection. 


Slope. 


Deflection. 


T o 


6W1 2 


4^73 


fl 


2fi* 


I . 


Ebh* 


Ebh* 


Eh 


ZEh 


,° 


2 WI* 


3 fT/3 

*Ebh* 


2fl 

3Ek 


iP 




Ebfc 


2 Eh 


3°- 


3 W 2 
4Ebfa 


1 Wl* 
4EbA* 


1 fl 

2 Eh 


(>Eh 


4°- 


2 Ebh* 


s JV/3 

32 Ebh* 


2fl 

3 Eh 


UEh 



So that, in the case of rectangular beams similarly loaded and 
supported, we may say that — 

Under a given load W, the slopes vary as the squares of 
the lengths, and inversely as the breadths and the cubes of the 
depths ; while the deflections vary as the cubes of the lengths, 
and inversely as the breadths and the cubes of the depths. 




Fig. 229. 



302 APPLIED MECHANICS. 

On the other hand, under their working-loads, the slopes vary 
directly as the lengths, and inverse]y as the depths ; while the 
deflections vary as the squares of the lengths, and inversely as 
the depths. 

§ 203. Beams Fixed at the Ends. — The only cases which 
we shall discuss here are the two following ; viz., — 

i°. Uniform section loaded at the middle. 

2°. Uniform section, load uniformly distributed. 

Case I. — Uniform Section loaded at tlie Middle. — The 
fixing at the ends may be effected by building the beam for 

some distance into the wall, as 
[" shown in Fig. 229. The same 
result, as far as the effect on 
the beam is concerned, might 
be effected as follows : Hav- 
ing merely supported it, and 
placed upon it the loads it has to bear, load the ends outside 
of the supports just enough to make the tangents at the sup- 
ports horizontal. 

These loads on the ends would, if the other load was re- 
moved, cause the beam to be convex upwards : and, moreover, 
the bending-moment due to this load would be of the same 
amount at all points between the supports ; i.e., a uniform 
bending-moment. Moreover, since the effect of the central 
load and the loads on the ends is to make the tangents over 
the supports horizontal, it follows that the upward slope at the 
support due to the uniform bending-moment above described 
must be just equal in amount to the downward slope due to the 
load at the middle, which occurs when the beam is only sup- 
ported. 

Hence the proper method of proceeding is as follows : — 

i°. Calculate the slope at the support as though the beam 
were supported, and not fixed, at the ends ; and we shall have, 
if we represent this slope by i„ the equation 



BEAMS FIXED AT THE ENDS. 303 



Wl- f , 

U =-^Ei (I) 

2°. Determine the uniform bending-moment which would 
produce this slope. 

To do this, we have, if we represent this uniform bending- 
moment by M„ that the slope which it would produce would be 

~^7 ; (2) 

and, since this is equal to i 1} we shall have the equation 

M£ _ Wfi ( v 

Wl 
■■ M x = -™ (4) 

This is the actual bending-moment at either fixed end ; and the 
bending-moment at any special section at a distance x from 
the origin will be 

where M is the bending-moment we should have at that sec- 
tion if the beam were merely supported, and not fixed. Hence, 
when it is fixed at the ends, we shall have, for the bending- 
moment at a distance x from O, where O is at the left-hand 
support, 

W W 
M=-x--l. (5) 

When x = -, we obtain, as bending-moment at the middle, 

Mo = — ; (6) 

and, since M x = — M oy it follows that the greatest bending- 
moment is 

Wl 

8' 



304 APPLIED MECHANICS. 

this being the magnitude of the bending-moment at the middle 
and also at the support. 

POINTS OF INFLECTION. 

The value of M becomes zero when 

x — - and when x = — : 
4 4 

hence it follows that at these points the beam is not bent, and 
that we thus have two points of inflection half-way between the 
middle and the supports. 

SLOPE AND DEFLECTION UNDER A GIVEN LOAD. 

We shall have, as before, 



CM , Wx* Wlx , 

\ = J Ei dx = ^7 - Jei + " 



and since, when x = o, i = o, 
.*. c — o 



dv W , _ . N /- v 

* = -r- = TTE-l 2 * ~ /X ) (7) 

dx 8£J 



U^ - ^), (8) 

\3 2 / 



JV/2X* 

S£/\J 



the constant vanishing because v = o when x = o. The slope 

becomes greatest when .r = -, and the deflection when x = -. 

4 2 

Hence for greatest slope and deflection, we have 

Wl 2 , V 

(9) 



64^/ 



*o = JTf ( IO ) 

I92ZSY 



BEAMS FIXED AT THE ENDS. 305 



SLOPE AND DEFLECTION UNDER THE WORKING-LOAD. 

If f represent the working-strength of the material per 

square inch, and if W represent the centre working-load, we 

shall have 

Wl ^fl 

8 ~ y 
/. W= 8 -f (11) 

8 is_y 24 Ey 

Case II. — Uniform Section, Load uniformly Distributed. — 
Pursuing a method entirely similar to that adopted in the former 
case, we have — 

i°. Slope at end, on the supposition of supported ends, is 

Wl* 



24^/ 

2 . Slope at end under uniform bending-moment M t is 



(1) 



2EI 
Hence, since their sum equals zero, 



<») 



mr Wl 

which is the bending-moment over either support. 
The bending-moment at distance x from one end is 

njr w ,i , Wl 

M=—(lx - x*) - — . (4) 

2/ 12 

This is greatest when # = 0, and is then . Hence great- 

12 
est bending-moment is, in magnitude, 

Wl , N 

(5) 

12 



306 applied mechanics. 

points of inflection. 
M becomes zero when x = - ± — 7 =. (6) 

2 2 y/3 

Hence the two points of inflection are situated at a distance 

— = on either side of the middle. 
2y/ 3 

SLOPE AND* DEFLECTION. 

/M W 

Ei dx= 7^EJ { ^- 2Xl - l * x) > (7) 

the constant vanishing because i = o when x = o. 

z> = — — i/x3 - — - — 1 (8) 

\2lEl\ 2 2 ) 

the constant vanishing because ^ = o when x = o. Hence for 
greatest slope and deflection we have, i is greatest when x = 

-I i ± -7= \ and v is greatest when ;tr = - ; 
2\ ^37 2 

v Q — . (10) 

SLOPE AND DEFLECTION UNDER WORKING-LOAD. 

For working-load we have 

m = /i 

12 y 



(«) 



, w = i£ (M) 






J—. 

zi-Ey 



(13) 



% = —Ar. (i4) 



BENDING-MOMENT AND SHEARING-FORCE. 



307 



EXAMPLES. 

i. Given a 4-inch by 12-inch yellow-pine beam, span 20 feet, fixed 
at the ends ; find its safe centre load, its safe uniformly distributed load, 
and its deflection under each load. Assume a modulus of rupture 5000 
lbs. per square inch, and factor of safety 4. Modulus of elasticity, 
1200000. 

2. Find the depth necessary that a 4-inch wide yellow-pine beam, 20 
feet span, fixed at the ends, may not deflect more than one four-hun- 
dredth of the span under a load of 5000 lbs. centre load. 



§ 204. Variation of Bending-Moment with Shearing- 
Force. — If, in any loaded beam whatever, M represent the 
bending-moment, and F the shearing-force at a distance x from 
the origin, then will 

(1) 



F= M 
ax 



Proof (a). — In the case of a cantilever (Fig. 230), assume 
the origin at the fixed end ; then, if M 
represent the bending-moment at a 
distance x from the origin, and M-\- AM 
that at a distance x -f- Ax from the 
origin, we shall have the following 
equations: — 

M= -S W{1- x), 



Fig. 230. 



x = l 



M 4- AM = -2 W{1 - x — Ax) nearly. 

X = X 

Hence, by subtraction, 



x = l 

AM = Ax*Z W nearly 



AX r =* 



308 APPLIED MECHANICS. 

and, if we pass to the limit, and observe that 



we shall obtain 



x = l 

F=$ IV, 



f=* w 



(b) In the case of a beam supported at the ends (Fig. 231), 
,_a_ assume the origin at the left-hand 

1 ^= 7s end, and let the left-hand support- 

ing-force be S ; then, if a represent 
Fi c 231. the distance from the origin to the 

point of application of W, we shall have the equations 






X = X 



M=Sx-% W(x - a), 



X = o 

X = X 



M + AM = S(x + A*) - 2 #%* - « + A*) nearly. 

X = o 

Hence, by subtraction, 

jr = jr 

A J/ = S . Ax — 2 WAx nearly 



. AM c '"' , 

.*. = o — S JT nearly : 

A* * = 



and, if we pass to the limit, and observe that 
F=S-i * IV, 



we shall obtain 

dM 



= F, (3) 

ax 



as before. 



\ 


\ / 


'J 

'a 



LONGITUDINAL SHEARING OF BEAMS. 309 

§ 205. Longitudinal Shearing of Beams. — The resistance 
of a beam to longitudinal shearing sometimes becomes a mat- 
ter of importance, especially in timber, where the resistance to 
shearing along the grain is very small. We will therefore pro- 
ceed to ascertain how to compute the intensity of the longi- 
tudinal shear at any point of the beam, under any given load ; 
as this should not be allowed to exceed a certain safe limit, to 
be determined experimentally. Assume a 
section AC (Fig. 232) at a distance x from 
the origin, and let the bending-moment at 
that section be M. Let the section BD be 
at a distance x -\- Ax from the origin, and 
let the bending-moment at that section be fig. 232. 

M+ AM. 

Let y Q be the distance of the outside fibre from the neutral 
axis ; and let ca = y z be the distance of a y the point at which 
the shearing-force is required, from the neutral axis. 

Consider the forces acting on the portion ABba, and we 
shall have — 

My 
i°. Intensity of direct stress at A = -j~. 

2°. Intensity of direct stress at a unit's distance from neu- 

, • M 
tral axis = -y. 

My 
3 . Intensity of direct stress at e, where ce = y, is -=-. 

(M -f AM)y 
So, likewise, intensity of direct stress at f is j . 

Therefore, if z represent the width of the beam at the point 
e, we shall have — 

Total stress on face Aa = ~y I yzdy, 

1 J yx 



Total stress on face Bo = j I yzdy 



310 APPLIED MECHANICS. 

,\ Difference = — — I yzdy : 

and this is the total horizontal force tending to slide the piece 
AabB on the face ab. 

Area of face ab, if z t is its width, is 

therefore intensity of shear at a is approximately 

AM fro 
~TJ n yzdy 

z^Ax y 
or exactly (by passing to the limit) 

tdM\ 



/am\ 
\~dx~) C y6 

~z-rL yzdy - 



And, observing that F = -j-, this intensity reduces to 

F C y ° 



-j yzdy. (i) 



We may reduce this expression to another form by observ- 
ing, that, if y 2 represent the distance from c to the centre of 
gravity of area Aa, and A represent its area, we have 

r*> 

I yzdy—y z A; 
*j yi 

therefore intensity of shear (at distance j x from neutral axis) at 
point a = 

jjcvo- (») 

This may be expressed as follows : — 



LONGITUDINAL SHEARING OF BEAMS. 311 

Divide the shearing-force at the section of the beam under 
consideration, by the product of the moment of inertia of the 
sectioji and its width at the point where the intensity of the 
shearing force is desired, and multiply the quotient by the statical 
moment of tJie portion of the cross-section between the point in 
question and the outer fibre ; this moment being taken about the 
neutral axis. The result is the required intensity of shear. 

The last factor is evidently greatest at the neutral axis ; 
hence the intensity of the shearing-force is greatest at the 
neutral axis. 

LONGITUDINAL SHEARING OF RECTANGULAR BEAMS. 

For rectangular beams, we have 

/=-, S I = *. 



Hence formula (2) becomes 



I2F 

M5(^)- (3) 



For the intensity at the neutral axis, we shall have, therefore, 

12F//1 bh\ _ 3 F 

~¥te V4 7/ 2 Jk' ^ 

since for the neutral axis we have 



h i bh 

- and A = — . 
4 2 



EXAMPLES. 



i. What is the intensity of the tendency to shear at the neutral axis 
of a rectangular 4-inch by 12-inch beam, of 14 feet span, loaded at the 
middle with 5000 lbs. 



312 



APPLIED MECHANICS. 



2. What is that of the same beam at the neutral axis of the cross- 
section at the support, when the beam has a uniformly distributed load 
of 12000 lbs. 

3. What is that of a 9-inch by 14-inch beam, 20 feet span, loaded 
with 15000 lbs. at the middle. 




§ 206. Strength of Hooks. — The following is the method 
to be pursued in determining the stresses in a 
hook due to a given load ; or, vice versa, the 
proper dimensions to use for a given load. 

Suppose (Fig. 233) a load hung at E ; the 
load being P, and the distance from EA to 
the inside of the most strained section being 

AB = n. 

Let O be the centre of gravity of this sec- 
tion, and let OB = y. Conceive two equal and 
opposite forces, each equal and parallel to P, 
acting at O. 

Let A = area of section, and let / = its 
moment of inertia about CD (BCDF represents the section 
revolved into the plane of the paper); then — 

i°. The downward force at O causes a uniformly distributed 
stress over the section, whose intensity is 

P 
* -7- 

2°. The downward force at E and the upward force at O 
constitute a couple, whose moment is 

P{n +y); 

and this is resisted, just as the bending-moment in a beam, by 
a uniformly varying stress, producing tension on the left, and 
compression on the right, of CD. 



SHORT STRUTS. 



313 



If we call p 2 the greatest intensity of the tension due to 
this bending-moment, viz., that at B, we have 



P(n + y)y . 

P2 = J , 



therefore the actual greatest intensity of the tension is 



and this must be kept within the working-strength if the load 
is to be a safe one. 



EXAMPLES. 

1. Suppose the hook to be made of i-inch diameter iron, and 
n = 1 inch : what is the working-load, modulus of rupture = 50000 
lbs. per square inch, factor of safety 6. 

2. A tension-rod hanging vertically bears a load at a horizontal dis- 
tance of three inches from its centre of gravity : find the necessary 
diameter, supposing it to be of wrought-iron. 

§ 207. Short Struts The case of a short strut, with the 

load applied at some point other than the 
centre of gravity of the section, is similar to 
that of the hook. Thus, let O' (Fig. 234) be 
the centre of gravity of the lower section, 
and let A'O = x Q . 

Conceive two equal and opposite forces 
at (7, each equal and parallel to P, and we 
have — 

i°. Downward force along line 00' causes uniform stress 
of intensity, 



Fig. 



P 

A 



2°. The other two form a couple, whose moment is 



314 APPLIED MECHANICS. 

therefore the greatest intensity of the compression due to this 

will be that at B, or 

{Px Q )a 

where a == O ' B '. Hence total greatest intensity is 

. P . Px Q a 

p = a + -t> 

or, if we write 

where p = radius of gyration of lower section about the axis 
through O' perpendicular to the plane of the paper, we have 



>=-l'+"i> 



and this should be kept within the limits of the working- 



strength of the material. 



EXAMPLES. 



i. Given a cylindrical column of 8 inches diameter, and let x Q = 2 
inches : find greatest stress per square inch under a load of 1 00000 lbs. 

2. Given P = 200000 lbs., x = 2 inches : find diameter of a 
yellow-pine strut suitable to bear the load, with factor of safety 4. 
Compressive strength of yellow pine == 4400 lbs. per square inch. 

§208. Strength of Columns. — The formulae in common 
use for the strength of columns are of three kinds ; viz, — 

i°. Euler's formulae, where it is assumed, that, for any given 
material, there is a certain definite ratio of length to diameter, 
below which a column will give way by direct crushing, while 
one whose ratio of length to diameter is greater will give way 
wholly by transverse bending. 

2°. Hodgkinson's empirical formulae, based upon his experi- 
ments upon small columns of a variety of ratios of length to 
diameter. 



GORDON'S FORMULM FOR COLUMNS. 315 

3 . Gordon's formulae, where it is assumed that all columns 
give way by a combination of crushing and bending. 

It is very much to be regretted that none of these sets of 
formulae are borne out by experiment upon the large scale, 
and that thus far we have no formulae for columns that are 
borne out generally. Euler's are evidently faulty in the funda- 
mental assumption ; Hodgkinson's experiments were made on 
small columns, and do not agree well with those on large ones ; 
Gordon's, or, as they are otherwise called, Rankine's, are prob- 
ably correct in their fundamental assumption, but there is a 
serious lapse in the reasoning by which they are deduced. 

The formulae most frequently used in American practice 
are those of Gordon. Hence we will take those first. 

§ 209. Gordon's Formulae for Columns. (a) Column 
fixed in Direction at Both Ends. — Let CAD be the cen- 
tral axis of the column, P the breaking-load, and v the 
greatest deflection, AB. Conceive at A two equal and 
opposite forces, each equal to P ; then — 

i°. The downward force at A causes a uniformly dis- 
tributed stress over the section, of intensity, 



Fig. 235. 



2°. The downward force at C and the upward force 
at A constitute a couple, whose moment is 

M = Pv; 

and this is resisted, just as the bending-moment in a beam, by 
a uniformly varying stress, producing compression on the right, 
and tension on the left, of A. 

If we call p 2 the greatest intensity of the compression due 
to this bending, we have 

+ _ Wy 

F* — * 7 



316 APPLIED MECHANICS. 

where y == distance from the neutral axis to the most strained 
fibre of the section at A. Then will the greatest intensity of 
the stress of compression at section A be 

P Pvy 
*=**+** 7* a + -p'' 

and, since Pis the breaking-load,/ must be equal to the break- 
ing-strength for compression per square inch =./. 
Hence 

'=$(-+?)-' <■> 

where p = smallest radius of gyration of section at A. 

Thus far the reasoning appears sound ; but in the next step 

it is assumed, that because, in a loaded beam, the greatest 

deflection under the breaking-load varies as the square of the 

length, and inversely as the distance from the neutral axis to 

the most strained fibre, therefore in this case it is assumed 

that we must have also 

I 2 





V oc — , 




or 


I / 2 

c y 




where c is 


a. constant to be determined 


sy 


therefore, 


substituting this in (i), 








I 




■ P- M 






I + — 
cp- 





Hence 



(2) 



GORDON'S FORMULA FOR COLUMNS. 



317 



which is the required formula for a column fixed in direction 
at both ends. 

(b) Column hinged at the Ends. — It is assumed in the 
previous case that the points of inflection are halfway 
between the middle and the ends, and hence that, by- 
taking the middle half, we have the case of bending of a 
column hinged at the ends (Fig. 236). Hence, to obtain 
the formula suitable for this case, substitute, in (2), 2/ 
for /, and we obtain fig. 236. 

M 



P = 



1 + 



4/f 



(3) 



(c) Column fixed at One End and hinged at the Other (Fig. 
237). — In this case we should, in accordance with these 
assumptions, take \ of the column fixed in direction at 
both ends ; hence, to obtain the formula for this case, 
substitute, in (2), \l for /, and we thus obtain 



fA 



1 + 



Fig. 237. 



i6/ 2 ' 



(4) 



Rankine gives, for values of f and c, the following, based 
upon Hodgkinson's experiments: — 



Wrought-iron 
Cast-iron . . 
Dry timber . 




36000 
6400 
3000 



3 l8 APPLIED MECHANICS. 

§ 210. Euler's Rules for the Strength of Columns. — The 

following are the rules for determining the strength of a col- 
umn of uniform cross-section, according to Euler : — 

(a) Column fixed in Direction at One End 07ily, which bends, 
as shown in the Figure. 

i°. Calculate the breaking-load on the assumption that the 
column will give way by direct compression. This will be 

where f = crushing-strength per square inch, and A = area 
of cross-section in square inches. 

2°. Calculate the load that would break the column if it 
were to give way by bending, by means of the following for- 
mula : — 



(5J 



where E = modulus of elasticity of the material, / =: smallest 
moment of inertia of the cross-section, and / = length of 
column. 

Then will the actual breaking-strength, according to Euler, 
be the smaller of these two results. 

To deduce the latter formula, assume the origin at the 
hinged end, and take x vertical and y horizontal. 

Let p = radius of curvature at point (x, y), and 
let M = bending-moment at the same point. 

Then we shall have, just as was shown in the 
case of the deflection of beams, 



M Py 



- p = Ti=ei' (3) 

But as was there shown, 

i d 2 y 

p ~ dx 2 



EULER'S RULES FOR STRENGTH OF COLUMNS. 319 



d 2 y _ P^ 
dx 2 ~ Ei y " 



,. -{ d l. d ^dx = ^[y d ldx 
J dx d 2 x El J dx 

' " \dx) 
and, since for 



~i/ + c; 



dy P , 



•*• 


dx y ei J 




dy fp 




Sa 2 - y 2 V EI 


.-. 


sin"^ = y^-.x + c. 
a V EI 


And since, when 




^ = 0, 


y = 0, 


and we have 





<T = O, 



-®-v/£ 



When 7 = a, we know that x = I ; hence, substituting in 
(5), we have 

7T M fP 



or 



2 ▼ 2£/ 



P-(5)'« (6) 



320 APPLIED MECHANICS. 

(b) Column hinged at Both Ends (Fig. 236). 

i°. Calculate the crushing-load, as before, from the formula 

P s =/A. 

•2°. Calculate the load that would break it, if it were to give 
way wholly by transverse bending, from the formula 



P* 



(tK (7) 



/ 

this being derived from (2) or (6) by substituting - for // the 

reasoning being the same for this substitution as was adopted 
with Gordon's formula. 

(c) Column fixed in Direction at Both Ends (Fig. 235). — 
We have for the crushing-load the same formula as before ; 
viz., — 

P*=fA; 

and for the bending we have 

/». = (?) V (8) 

/ 
this being obtained from (2) or (6) by substituting - for /. 

(d) These rules may be summed up as follows : — 
i°. Calculate the crushing-load by the formula 

Pi = /A. 



2 . Calculate the load that would break the column by 
ding, from the following formulae : — 



EI 



HODGKINSON'S RULES FOR STRENGTH OF COLUMNS. 32 1 

if fixed in direction at one end only ; 
if hinged or rounded at both ends ; 

(r) * = (=)'*/ 

if fixed in direction at both ends. 

Then will the actual breaking-strength be the least of the 
two results. 

(e) In order to ascertain the length where incipient flexure 
occurs, according to this theory we should place the two results 
equal to each other, and from the resulting equation determine /. 
We should thus obtain, for the three cases respectively, — 

w /=yg, (9) 

(« '~Vfr (lo) 

(y) /=a Vff (II > 

Hence all columns whose length is less than that given hr 
these formulae will, according to Euler, give way by direct 
crushing ; and those of greater length, by bending only. 

§ 211. Hodgkinson's Rules for the Strength of Columns. 
— Eaton Hodgkinson made a very extensive series of tests of 
columns, especially of cast-iron, and deduced from these tests 
certain empirical formulae. These tests form, even at the pres- 
ent time, the basis of the most used formulae for the strength 
of columns. The strength of pillars of the ordinary sizes used 
in practice has been computed by means of Hodgkinson's for- 



322 APPLIED MECHANICS. 

mulae, and tabulated by Mr. James B. Francis : and we find in 
his book the following rules for the strength of solid cylindrical 
pillars of cast-iron, with the ends flat; i,e., "finished in planes 
perpendicular to the axis, the weight being uniformly distrib- 
uted on these planes." 

For pillars whose length exceeds thirty times their diameter, 
he gives the formula, 

W= 99318 £?, (!) 

where D = diameter of column in inches, / = length in feet, 
W =■ breaking-weight in pounds. 

If, on the other hand, the length does not exceed thirty times 
the diameter, he gives, for the breaking-weight, the following 
formula : — 

W = WC , (2) 

where W = breaking-weight that would be derived from the 
preceding formula, W = actual breaking-weight, c = weight 
which would crush the pillar, or 

c = io 9 8oir A ( 3 ) 

For hollow cast-iron pillars, if D = external diameter in inches, 
d = internal diameter in inches, we should have, in place of (1), 

^ = 99318 — , (4) 

and in place of (3), 

c — 109801 — i <-. (5) 

4 

For very long wrought-iron pillars, Hodgkinson found the 
strength to be 1.745 times that of a cast-iron pillar of the same 
dimensions ; but, for very short pillars, he found the strength of 



=2-' 



STRENGTH OF SHAFTING. 323 

the wrought-iron pillar very much less than that of the cast-iron 
one of the same dimensions. With a length of 30 diameters 
and flat ends, the wrought-iron exceeded the cast-iron by about 
ten per cent. 

§ 212. Strength of Shafting. — The usual criterion for the 
strength of shafting is, that it shall be sufficiently strong to 
resist the twisting to which it is exposed in the transmission of 
power. 

Proceeding in this way, let EF (Fig. 239) be a shaft, AB the 
driving, and CD the following, pulley. 
Then, if two cross-sections be taken 
between these two pulleys, the por- 
tion of the shaft between these two E u 
cross-sections will, during the trans- 
mission of power, be in a twisted con- 

1 ' Fig. 239. 

dition ; and if, when the shaft is at 

rest, a pair of vertical parallel diameters be drawn in these sec- 
tions, they will, after it is set in motion, no longer be parallel, 
but will be inclined to each other at an angle depending upon 
the power applied. Let GH be a section at a distance x from 
% and let KI be another section at a distance x -f- dx from O. 
Then, if di represent the angle at which the originally parallel 
diameters of these sections diverge from each other, and if r = 
the radius of the shaft, we shall have, for the length of an arc 
passed over by a point on the outside, 

rdi; 

and for the length of an arc that would be passed over if the 
sections were a unit's distance apart, instead of dx apart, 

rdi _ di 
dx dx 

This is called the strain of the outer fibres of the shaft, as it 
is the distortion per unit of length of the shaft. 



324 APPLIED MECHANICS. 

In all cases where the shaft is homogeneous and symmet- 
rical, if i is the angle of divergence of two originally parallel 
diameters whose distance apart is x, we shall have the strain, 

di i 
v = r — > = r-. 

dx x 

This also is the tangent of the angle of twist. 

A fibre whose distance from the axis of the shaft is unity, 
will have, for its strain, 

di i 
dx x 

A fibre whose distance from the axis of the shaft is p, will have, 

for its strain, 

di i 

v = p— = p-. 

dx x 

Fixing, now, our attention upon one cross-section, GH, we have 
that the strain of a fibre at a distance p from the axis (p varying, 
and being the radius of any point whatever) is 



<©• 



1 
where - is a constant for all points of this cross-section. 

Hence, assuming Hooke's law, " Ut tcnsio sic vis," we shall 
have, if C represent the shearing modulus of elasticity, that the 
stress of a fibre whose distance from the axis is />, is 



>-»-<Ks)-«® 



which quantity is proportional to p, or varies uniformly from the 
centre of the shaft. 

The intensity at a unit's distance from the axis is 



© 



STRENGTH OF SHAFTING. 



325 




Fig. 240. 



and if we represent this by a, we shall have for that at a dis- 
tance p from the axis, 

p = ap. 

Hence we shall have (Fig, 240), that, on a small 
area, 

dA = dp(pdO) — pdpdO, 

the stress will be 

pdA = apdA — ap 2 dpd$. 

The moment of this stress about the axis of the shaft is 

ppdA = ap 2 dA = ap^dpdO, 

and the entire moment of the stress at a cross-section is 

afp 2 dA — affpidpdO = al, 

where / = fp 2 dA is the moment of inertia of the section about 
the axis of the shaft. 

This moment of the stress is evidently caused by, and hence 
must be balanced by, the twisting-moment due to the pull of the 
belt. Hence, if M represent the greatest allowable twisting- 
moment, and a the greatest allowable intensity of the stress at 
a unit's distance from the axis, we shall have 

M = a J =-J. 
P 

If / is the safe working shearing-strength of the material 
per square inch, we shall have / as the greatest safe stress per 
square inch at the outside fibre, and hence 

M=-I 
r 

will be the greatest allowable twisting-moment. 



326 APPLIED MECHANICS. 



For a circle, radius r, 

2 y 2 y 16 

For a hollow circle, outside radius r„ inside radius r 2 , 

7r(r, 4 — r 2 4 ) 7r 

1= — — .\ jlf = / — (r x 4 - r a 4). 

Moreover, if the dimensions of a shaft are given, and the 

actual twisting-moment to which it is subjected, the stress at a 

fibre at a distance p from the axis will be found by means of the 

formula 

Mp 

The more usual data are the horse-power transmitted and 
the speed, rather than the twisting-moment. 

If we let P =. force applied, and R — its leverage, as, for 
instance, when P = difference of tensions of belt, and R = 
radius of pulley, we have 

M= P.R; 

and if HP = number of horses -power transmitted, and 
N = number of turns per minute, then 

HP= — ' 

33000 

2irN 



EXAMPLE. 

Given working-strength for shearing of wrought-iron as 10000 lbs. 
per square inch; find proper diameter of shaft to transmit 20-horse 
power, making 100 turns per minute. 



TRANSVERSE DEFLECTION OF SHAFTS. 327 



§213. Angle of Torsion. — From the formula, § 212, 
/ 



p = — -, combined with 



we have 



p = a P = Cp-, 

x 

i _ Mp 

' T 

Mx 



Cp 

x I 



■■ ' = ci' 

which gives the circular measure of the angle of divergence of 
two originally parallel diameters whose distance apart is x ; the 
twisting-moment being M, and the modulus of shearing elas- 
ticity of the material, C. 

EXAMPLES. 

1. Find the angle of twist of the shaft given in example 1, § 212, 
when the length is 10 feet, and C = 8500000. 

2. What must be the diameter of a shaft to carry 80 horses-power, 
with a speed of 300 revolutions per minute, and factor of safety 6, break- 
ing shearing-strength of the iron per square inch being 50000 lbs. 

§ 214. Transverse Deflection of Shafts. — In determining 
the proper diameter of shaft to be used in any given case, we 
ought not merely to consider the resistance to twisting, but 
also the deflection under the transverse load of the belt-pulls, 
weights of pulleys, etc. This deflection should not be allowed 
to exceed yj-g- of an inch per foot of length. Hence the de- 
flection should be determined in each case. 

The formulae for computing this deflection will not be given 
here, as the methods to be pursued are just the same as in the 
case of a beam, and can be obtained from the discussions on 
that subject. 



328 APPLIED MECHANICS. 

§ 215. Combined Twisting and Bending. — The most com- 
mon case of a shaft is for it to be subjected to combined twisting 
and bending. The discussion of this case involves the theory 
of elasticity, and will not be treated here ; but the formulae com- 
monly given will be stated, without attempt to prove them until 
a later period. These formulae are as follows : — 
Let M x = greatest bending-moment, 
M 2 = greatest twisting-moment, 
r = external radius of shaft, 
/ = moment of inertia of section about a diameter, 

for a solid shaft / = , 

4 
f = working-strength of the material = greatest al- 
lowable stress at outside fibre ; 



then 



According to Grashof, 



L s 



/= j\W* + t^« a + ^ 2 j- (0 



2°. According to Rankine, 



f=?\M l +)/M l * + M 2 *\. (2) 



GENERAL REMARKS. 329 



CHAPTER VIL 

STRENGTH OF MATERIALS AS DETERMINED BY 
EXPERIMENT. 

§216. General Remarks. — Whatever computations are 
made to determine the form and dimensions of pieces that 
are to resist stress and strain, must, if they are to have any 
practical value, be based upon experiments made upon the ma- 
terials themselves. 

The most valuable experiments in any given case, whenever 
the results of such experiments are available, are those made 
upon pieces of the same quality, size, and form as those to 
which the results are to be applied, and under conditions en- 
tirely similar to those to which the pieces are subjected in 
actual practice. 

It is very seldom, that the results of such experiments are 
available ; and hence we must, in general, make use of such 
tests as have been or can be made, and from them determine 
the strength of the pieces in actual use by computation, making 
good use of our judgment. 

As time goes on, and experimental science advances, a 
greater number of the conditions that exist in actual practice 
are introduced into the experiments ; and hence the reliability 
of the experimental results, and their applicability to practical 
cases, are increased. Nevertheless, it is necessary to use the 
utmost caution when applying the results of the experiments to 
cases where the conditions are different from those under which 
the experiments were made. An attempt will be made in this 



330 APPLIED MECHANICS. 

chapter to give an account of the most important results of 
experiment on the strength of materials, and to explain the 
modes of using the results that are now employed. As to 
the way in which the experiments have generally been carried 
on, we may observe : — 

i°. In by far the greater number of cases, the test pieces have 
been very much smaller than the pieces to be used. Indeed, 
it is only of late years that the importance of testing full-size 
pieces has been recognized ; and hence most of the experiments 
upon such pieces are of very recent date. 

2°. In the greater part of the experiments that have been 
made, the phenomena observed have been those that occur 
during the application of the load for a short time only ; very 
little having been done by way of determining the behavior of 
the pieces under a long-continued action of the load, or under 
repeated applications of the load, such as occur in practice. 

3°. Very few experiments have been made on the effect of 
applying two kinds of stress simultaneously, as tension and 
bending, or twisting and bending, or on applying stresses of 
opposite kinds, as tension and compression, successively. 

4°. The tests thus far made have had for their object more 
frequently to determine the breaking-strength of the piece. 
Next to this, the subject most frequently experimented upon 
has been the limit of elasticity ; and less has been done by way 
of determining the modulus of elasticity, and other matters. 

5°. The fact that the breaking-strength alone is not a suffi- 
cient criterion by which to determine the suitability of a mate- 
rial for use in construction, has been recognized only by the 
later experimenters. 

6°. In order to understand what is meant by " the limit of 
elasticity," we must observe, that, if a small load be applied to 
the piece under test, and then removed, the deformation or dis- 
tortion caused by the application of the load apparently van- 
ishes, and the piece resumes its original form and dimensions 



GENERAL REMARKS. 33 I 

on the removal of the load ; in other words, no permanent set 
takes place. When the load, however, is increased beyond a 
certain point, the piece under test does not return entirely to 
its original dimensions on the removal of the load, but retains a 
certain permanent set. 

The load upon the application of which permanent set 
apparently begins, is called the limit of elasticity, and is found 
by experiment to be at about one-third the breaking-weight in 
iron, and from one-third upwards in steel, sometimes reaching 
nearly three-fourths. 

Experiments show, however, that even a very small load will 
produce a permanent set, and that the apparent return of the 
piece to its original dimensions upon the removal of the load is 
only due to the want of delicacy in the measuring-instruments 
that have been used in the tests. A better definition of the 
limit of elasticity would therefore be, that load upon the appli- 
cation of which the permanent set begins to be noticeable, with 
such rough means of measuring as a pair of dividers. 

7°. It has often been assumed, that, if the load applied to 
the piece in practice exceeded the elastic limit, the piece would 
be permanently injured in its properties for resisting stress, and 
that the deformation and injury would continue increasing, until 
eventually fracture would occur. It has been proved experimen- 
tally, however, that it is sometimes advantageous to apply once 
a load to a piece somewhat greater than the elastic limit, and 
that by this means the elastic limit is increased. This process 
of using up a part of the elasticity of the piece cannot continue 
indefinitely, and the data to show how far it can be advanta- 
geously carried are but few. 

8°. The determination of the modulus of elasticity, which 
has been defined (§ 167) as the ratio of the stress to the strain, 
is a very important matter ; as it gives us the means of com- 
puting the deformation under any given load, and thus deter- 
mining the safe load by prescribing the greatest deformation to 



332 APPLIED MECHANICS. 

be allowed, rather than by prescribing that the safe load shall 
be a certain fraction of the breaking-load. 

9°. Other important matters which guide us in judging of 
the suitability of a piece for the purpose to which it is to be 
applied, are, the appearance of its fracture, its density, its homo- 
geneity, its composition, and the care taken in its manufacture, 
or the circumstances of its growth and seasoning if it is wood, 
also its brittleness, hardness, malleability, ductility, the amount 
of warning it gives before giving way, etc. 

io°. From such data as we have furnished to us by experi- 
ment, we decide, to the best of our ability, as to the suitability 
of the piece for the use for which it is intended, and also as to 
the amount it will safely bear, when we know its dimensions, 
or the proper dimensions to bear safely the required stress. 

u°. Tests have been made on tension, compression, shear- 
ing, transverse and torsional strength, of the different materials 
used in construction, especially cast-iron, wrought-iron, steel, 
and wood, also copper and other metals ; but the tests on ten- 
sile strength are by far the most numerous in the case of iron. 

§ 217. Cast-iron. — Cast-iron is a combination of iron with 
2 per cent to 6 per cent of carbon. The large amount of carbon 
which it contains is its distinguishing feature, and determines 
its behavior in most respects. 

Pig-Iron is the result of the first smelting, being obtained 
directly from the smelting-furnace. The ore and fuel are put 
into the furnace, together with a flux, which is of a calcareous 
nature when the ore is argillaceous, or which contains clay 
when the ore is calcareous. The mass is brought to a high 
heat, a strong blast of air being introduced. The mass is thus 
melted ; the fluid iron settling to the bottom, while slag, which 
is the result of the combination of the flux with the impurities 
of the ore, rises to the top. The iron is drawn off in the liquid 
state, and run into moulds, the result being pig-iron. 

The result of this first melting is very rarely used for any 



CAST-IRON. 33, 



casting; but the pig-iron is usually re-melted in a cupola furnace 
before being used, the result of this re-melting being the ordi- 
nary cast-iron of commerce. 

The pig-iron is divided into classes, according to the purpose 
for which it is intended, and the amount of carbon it contains. 

Those pigs that have a considerable amount of carbon in me- 
chanical mixture, and show a gray color on being fractured, are 
used by the founder to melt over, and make cast-iron. This is 
called "foundry iron," and is divided into foundry iron Nos. I, 
2, and 3, from which are subsequently made gray cast-iron 
Xos. I, 2, and 3. When there is less carbon, it is sometimes 
called " foundry No. 4," etc. ; but it is only used to make 
wrought-iron of an inferior quality. 

Those pigs which are to be used in making wrought-iron 
and steel, and which have been fused at a low heat and with 
little fuel, are called u forge-iron." 

Cast-iron is of two kinds, white cast-iron and gray cast-iron. 
The first is a chemical compound of iron with 2 per cent to 6 per 
cent of carbon, almost all of the carbon being chemically com- 
bined with the iron. The second, or gray cast-iron, contains 
part of the carbon in chemical combination, and the remainder 
in the state of graphite mechanically mixed with the iron. 

Gray Cast-Iron is divided into three classes, known respec- 
tively as Nos. 1, 2, and 3. 

No. I contains the largest amount of carbon in mechanical 
mixture, the effect of which is to render it soft and fusible, 
though not as strong as Nos. 2 and 3. It is, therefore, very 
suitable for making castings where precision in form is a desid- 
eratum, as its fusibility causes it to fill the mould well. It is 
not as suitable, however, where strength is required. 

No. 2 is that which is most suitable for use in construction, 
as it is stronger than No. 1, and not so soft. 

No. J, on the other hand, contains the smallest amount of 
carbon in the graphitic form, and is, hence, harder and more 



334 APPLIED MECHANICS. 

brittle. It is suited, therefore, only for the massive and heavy 
parts of machinery. 

White Cast-Iron contains hardly any free carbon. It is of two 
kinds, granular and crystalline. The crystalline variety is of 
no use in construction : it is hard and very brittle. The granu- 
lar variety is also unsuitable for use in construction, but forms 
the hard skin on the surface of a piece that has been chilled. 

As to the adaptability of cast-iron to construction, it pre- 
sents certain advantages and certain disadvantages. It is the 
cheapest form of iron. It is easy to give it any desired form. 
It resists oxidation better than either wrought-iron or steel. It 
has a very high compressive strength. On the other hand, its 
tensile strength is comparatively small, averaging, in common 
varieties, 15000 pounds per square inch, or thereabouts. It 
cannot be riveted or welded when broken. It is brittle, break- 
ing off without giving much warning, and stretching but little 
before giving way. It is liable to hidden and small surface 
defects and air-bubbles, which render its strength somewhat 
doubtful. It is also liable to absorb impurities from the fuel or 
flux in the furnace, the most injurious being sulphur and phos- 
phorus ; the effect of the former being to produce red- short- 
ness, or brittleness when hot, and that of the latter to produce 
cold shortness, or brittleness when cold. 

Another very serious drawback in the use of cast-iron in con- 
struction is its liability to initial strains from the inequality 
in cooling. Thus, if one part of the casting is very thin and 
another very thick, the thin part cools first, and, in cooling, con- 
tracts ; and the thick part, cooling afterwards, causes stresses 
in the thin part, which may be sufficient to break it, or, if not, 
there may be so much stress established, that but little more 
will break it. Thus, the change of temperature from summer 
to winter is sometimes sufficient to break the arms of a pulley 
from off the rim. Its quality depends largely upon its composi- 
tion and its density. 



CAST-IRON. 335 



The fracture should be of a bluish-gray color, and close- 
grained texture, with considerable metallic lustre if the iron is 
of good quality. If the fracture is mottled, with patches of 
darker or lighter iron, or crystalline spots, it is an indication 
of unsoundness, especially so if there are air-bubbles. 

It is not well adapted to bear tension, on account of its low 
tensile strength, and also on account of its brittleness and 
treacherousness. 

In former times it was extensively used for iron beams to 
bear a transverse load, but has now been almost entirely super- 
seded by wrought-iron in this regard. 

It is still used for columns and posts in buildings, on ac- 
count of its high compressive strength ; but, in cases where the 
length of the column is so great as to cause it to give way by 
bending, as in bridge columns, its use has been almost wholly 
abandoned, and wrought-iron and steel are taking its place. In 
the case of bridge columns, it is also necessary to use a metal 
which can easily be riveted, and to which the other members 
can be readily attached; and wrought iron is more suitable than 
cast for this purpose : also another reason is, that wrought-iron 
and steel are much better suited to resist shocks than cast-iron. 

In machinery, it is used in all those parts where weight, 
mass, or form is of more importance than strength, as in the 
frames and bed-plates of machines, also for hangers, pulleys, 
and gear-wheels. 

Cast-iron is also used for water-mains where great pressure 
is to be resisted, also in hydraulic presses, and in heavy ord- 
nance. For shafting, wrought-iron has taken its place, and so 
also for the shells of steam-boilers, with the exception of some 
sectional boilers, partly on account of its low tensile strength 
and general untrustworthiness, but especially because of its 
liability to give way without warning when subjected to the 
sudden expansions and contractions which it would have to 
undergo if used in a steam-boiler. 



33^ APPLIED MECHANICS. 

Malleable Cast-iron. — -When a casting is to be made in a 
rather intricate form, it is frequently the custom to malleableize 
the cast-iron. This is done by heating it to a bright-red heat in 
an annealing oven, in powdered hematite ore, with a suitable 
flux. By this process a part of the carbon is removed, and the 
result is — provided the casting is not large — a product that 
can be hammered into any desired shape when cold, but is very 
brittle when hot. It is used in cases where toughness is re- 
quired, together with the possession of an intricate form : thus 
gun-locks, pokers, tongs, etc., are sometimes made by this 
means, and sometimes also screw propellers. 

§ 218. Tensile and Compressive Strength of Cast-Iron. 
— A list of the principal experimenters on the strength and 
elasticity of cast-iron will be given, and references to the 
accounts of their work, which the student who wishes to pursue 
the subject further will do well to consult : — 

i°. Eaton Hodgkinson : (a) Report of the Commissioners on the 

Application of Iron to Railway Structures. 
{I?) London Philosophical Transactions. 1840. 
(c) Experimental Researches on the Strength and other Properties 

of Cast-iron. 1846. 
2 . W. H. Barlow : Barlow's Strength of Materials. 
3 . Sir William Fairbairn : On the Application of Cast and Wrought 

Iron to Building Purposes. 
4 . Major Wade (U.S.A.) : Report of the Ordnance Department on the 

Experiments on Metals for Cannon. 1856. 
5 . Capt. T. J. Rodman : Experiments on Metals for Cannon. 
6°. Col. Rosset : Resistenza dei Principali Metalli da Bocchi di Fuoco. 
7 . John Anderson : Strength of Materials. 

Taking up first the experiments of Hodgkinson. and the 
other members of the commission, we find that they consist 
of a series of tests made to compare the strength of iron from 
different parts of the kingdom. These tests show the average 
strength of English cast-iron at that time. 



TENSILE, ETC., STRENGTH OF CAST-IRON. 



337 



The following table gives the tensile and also the crushing 
strength per square inch of the different kinds tested : — 





Tensile Strength 


Height of 


Crushing-Strength 


Description of the Iron. 


per 

Square Inch, 

in lbs. 


Specimen, in 
inches. 


per Square Inch, 
in lbs. 


Lowmoor, No. i 


I2694 


M 


64534 
56445 


Lowmoor, No. 2 


15458 


M 


995 2 S 
92332 


Clyde, No. 1 


16125 


1.1 


92896 

88741 


Clyde, No. 2 


17807 


1.1 


IO9992 
IO203O 


Clyde, No. 3 


23468 


u 


IO7197 
I 0488 I 


Blaenavon, No. 1 


13938 


u 


90860 

80561 


Blaenavon, No. 2 (first sample) 


16724 


14 


I 17605 
IO2408 


Blaenavon, No. 2 (second sample), 


I429I 


14 


68559 
68532 


Calder, No. 1 


J3735 




72193 
75983 


Coltness, No. 3 


15278 


14 


IOOI80 
IO1831 


Brymbo, No. 1 


14426 


14 


74815 
75678 


Brymbo, No. 3 


15508 


14 


76133 
76958 


Bowling, No. 2 


I35 11 


14 


76132 

739 8 4 


Ystalyfera, No. 2 (anthracite) . . 


145" 


14 


99926 
95559 


Yniscedwyn, No. 1 (anthracite) 


I395 2 


14 


83509 
78659 


Yniscedwyn, No. 2 (anthracite) 


13348 


14 


77124 
75369 


Morries Stirling's (second quality), 


25764 


14 


125333 
1 19457 


Morries Stirling's (third quality) . 


23461 


14 


158653 
129876 



35o APPLIED MECHANICS. 

The specimens used for determining the tensile strength 
were all cruciform in section ; those for determining the crush- 
ing-strength were cylinders, with a diameter of f inch. The 
machine used was simply a single lever, with a scale-pan at the 
extreme end, in which weights were placed. There was no 
provision for taking up the stretch of the specimen, and there- 
fore the specimens were all short. 

The average tensile strength of these specimens, omitting 
the Stirling metal, was 15298 lbs. per square inch; and the 
average compressive strength was 82296 lbs. per square inch. 

It will be observed, that, omitting the Stirling iron, which 
is a mixture of cast and wrought iron, the tensile strength 
ranged from 12694 to 23468 lbs. per square inch, and the com- 
pressive strength, from 56445 to 11 7605 lbs. per square inch. 
The reader will doubtless observe, that, as a rule, the crushing- 
strengths obtained from the longer specimens were less than 
those obtained with the shorter ones ; but this, it seems to the 
writer, is probably due to the nature of the testing-machine, 
and not to any bending in the specimen due to its length. 

Mr. Hodgkinson next proceeded to compare the tensile 
strength of specimens cruciform in section, with that of speci- 
mens circular in section and of about the same area. He found 
but little difference ; and this could readily be accounted for by 
the fact, that, the perimeter of the cruciform one being greater, 
the proportion of hard skin would be greater in the cruciform 
than in the circular, the effect of this being, perhaps, partially 
counteracted by some little initial stress, on account of unequal 
cooling of the different parts. 

Hodgkinson also made a few experiments to determine the 
laws of extension of cast-iron, and for this purpose used rods 
10 feet long and 1 square inch in section. The table of average 
results is the following : — 



RESULTS OF TESTS. 



339 



RESULTS OF NINE TENSILE TESTS. 



Weights laid 


Extensions, 




Strains, in 


Sets, in 


Ratio of 


Modulus 


on, 


in 


Sets, 


fractions of 


fractions of 


Weight to 


of 


in lbs. 


inches. 


in inches. 


the length. 


the length. 


Extensions 


Elasticity. 


105377 


O.OO90 


- 


O.OOOO7 


- 


I 17086 


I4050320 


1580.65 


O.OI37 


O.O0O22 


O.OOO 1 1 


O.OOOOOI8 


H5I3I 


138 1 5720 


2107.54 


O.OI86 


O.OOO55 


O.OOOI6 


O.OOOOO46 


I 13309 


13597080 


3161.3I 


O.0287 


O.OOIO7 


O.OOO24 


O.OOOO089 


IIOI50 


I321800O 


4215.08 


O.O39I 


O.OOI75 


O.OOO33 


O.OOOOI46 


IO7803 


12936360 


5268.85 


O.050O 


O.OO265 


O.OOO42 


O.000O22I 


105377 


I2645240 


6322.62 


O.0613 


O.OO372 


O.OOO51 


O.OOOO310 


IO3142 


I2377040 


7376-39 


O.O734 


O.OO517 


O.OO061 


O.OOOO431 


IOO496 


12059520 


8430.16 


O.0859 


O.O0664 


O.OOO72 


O.OOO0553 


9^39 


1 1776680 


9483.94 


O.0995 


O.O0844 


O.OO083 


O.OCOO703 


953i6 


I H37920 


IO537.7I 


O.1 136 


O.OI062 


O.OOO95 


O.OOO0885 


92762 


III3144O 


1 1 591.48 


O.I283 


O.OI306 


O.OOIO7 


0,0001088 


90347 


I 084 I 64O 


12645.25 


O.I448 


O.O1609 


O.OOI2 1 


O.OOO 1 34 1 


87329 


IO479480 


13699.83 


0.1668 


O.02097 


O.OOI39 


O.OOOI748 


82133 


9855960 


I4793.IO 


O.1859 


O.O24IO 


O.OOI55 


O.0002008 


79576 


9549120 



RESULTS OF EIGHT COMPRESSIVE TESTS. 



Weights laid 

on, 

in lbs. 


Compres- 
sions, 
in inches. 


Sets, 
in inches. 


Strains, in 
fractions of 
the length. 


Sets, in 
fractions of 
the length. 


Ratio of 
Weight to 
Compres- 
sions. 


Modulus 

of 
Elasticity. 


2064.75 


O.OI875 


O.OOO47 


O.OO0l6 


O.OOOOO39 


IIOI20 


I 32 I 4400 


4129.49 


O.03878 


O.OO226 


O.OOO32 


O.OOOO188 


I06485 


127782OO 


6194.24 


O.05978 


O.OO40O 


O.OOO5O 


O.OOOO333 


IO3617 


12434040 


8258.98 


O.07879 


O.O0645 


O.00066 


O.OOOO538 


IO4823 


I2578760 


IO323.73 


O.09944 


O.O0847 


O.OO083 


O.OOOO706 


IO3819 


I2458280 


I2388.48 


O.I2030 


O.OI088 


O.OO IOO 


O.OOOO907 


IO2980 


I235760O 


14453-22 


0.1 41 63 


O.OI405 


O.OOI18 


O.OOOII7I 


102049 


I2245880 


16517.97 


O.16338 


O.OI7I2 


O.OOI36 


O.OOOI427 


IOII02 


12132240 


18582.71 


O.18505 


O.02051 


O.OO I 54 


O.OOOI709 


IOO420 


I20504OO 


20647.46 


O.2C624 


0.02484 


O.OOI72 


O.OOO2070 


IOOII4 


I 2013680 


24776.95 


O.24961 


O.03220 


O.OO208 


O.OOO2683 


99263 


1 191 1 560 


28906.45 


O.29699 


O.0430O 


O.OO247 


OOOO3583 


9733 1 


H679720 


33030.80 


0.35341 


O.06096 


O.OO295 


O.OOO5080 


93463 


II215560 



340 



APPLIED MECHANICS. 



These tables show that the modulus of elasticity of cast- 
iron varies with the load, growing gradually smaller as the load 
increases. 

The following table enables us to compare the modulus of 
elasticity for tension with that for compression under nearly the 
same load. 



Tension. 


Compression. 


Load 

per Square 

Inch. 


Modulus of 
Elasticity. 


Load 

per Square 

Inch. 


Modulus of 
Elasticity. 


2107 

4215 
6322 
8430 

10537 
12645 

14793 


13597080 
12936360 
12377040 
11776680 
11131440 
10479480 
9549120 


2064 

4129 

6194 

8258 

10323 

12388 

*4453 


13214400 
12778200 
12434OOO 
12578760 
12458280 
12357600 
12245880 



This table shows, that, with moderate loads, the modulus of 
elasticity for tension of cast-iron does not differ materially from 
that for compression, and that the difference increases as the 
load becomes greater. 

This fact is one of very considerable importance, inasmuch 
as it is one of the fundamental assumptions in the common 
theory of beams ; and we thus see that experiment justifies our 
making this assumption for cast-iron whenever the load is not 
excessive. The justification is even greater with wrought-iron 
and steel. 

The gradual decrease of the modulus of elasticity with the 
increase of load shows that Hooke's law, " Ut tcnsio sic vis " 
(the stress is proportional to the strain), does not hold true in 



TENSILE, ETC., STRENGTH OF CAST-IRON. 



341 



cast-iron ; nevertheless, it is more nearly true for moderate loads 
than for larger ones. 

The experiments of Major Wade of the United-States army, 
most of which were made at the South-Boston Iron Foundry, 
are recorded in the Report of the Ordnance Department on 
metals for cannon, printed in 1856. The Report contains 
experiments in regard to the effect of keeping the iron in a 
state of fusion for a long time, and also on the effect of suc- 
cessive re-meltings upon the quality of the iron ; and he says, — 

" It is found that the same iron is greatly improved in qual- 
ity by retaining it in the furnace after it is melted for consider- 
able periods of time, and that it is further improved by casting 
it into pigs, and again re-melting it. But it is known also that a 
continuation of this process will ultimately impair the tenacity 
of the iron, and render it wholly unfit for use. It is found also 
that the different kinds of iron require a different kind of treat- 
ment to produce the best effect. The breaking-instrument en- 
ables one to ascertain the effect produced by these processes in 
all their several stages of progress, and to decide on that which 
is found most suitable for making guns of the best quality." 

On p. 279 of the Report is given the following experiment 
in this regard : — 

" The first sample was a rough, 
crude pig of grade No. 1, Green- 
wood iron. 

" The second, third, and fourth 
are the same iron, cast and cooled 
in like manner, and differ only in 
the number of times melted ; and 
they exhibit the changes effected 
in the strength by the repeated 
meltings only. The fifth sample 
is from the same melting as the fourth, from which it differs 
only in cooling, being cast in a large mass, and cooled slowly." 





Fusion. 


Tensile 
Strength. 


Pig . 


I St 


14000 




r 2d 


20900 


Bars . 


J3d 


30229 




(4th 


35786 


Head, 


4th 


337 2 4 



342 



APPLIED MECHANICS. 



The compressive strength of cast-iron tested by Major 
Wade varied from 84500 lbs. per square inch to 175000 lbs. per 
square inch. The following table of results of a number of 
tests of metal taken from different cannon, gives the average 
tensile strength, specific gravity, and proportion of carbon in 
the lot of specimens examined : — 







Tensile 










Specific 
Gravity. 


Strength, 

lbs., 

per Square 

Inch. 


Total 
Carbon. 


Combined 
Carbon. 


Allotropic 
Carbon. 


ist-class guns, 


7.204 


28805 


O.0384 


O.0178 


0.0206 


2d-class guns, 


7-154 


24767 


0.0376 


0.0146 


0.0230 


3d-class guns, 


7.087 


20148 


O.0365 


0.0082 


0.0283 



It will be noticed that an increase in tensile strength is 
generally accompanied by an increase in specific gravity. 
Thus another lot of iron gave the following results : — 





Tensile 

Strength, 

lbs., per 

Square Inch. 


Specific 
Gravity. 


Mean . 

Least 

Greatest, 


27232 
22402 
31027 


7.302 
7-I63 

7.402 



He also experimented on the difference between hot and 
cold blast iron, and recommends decidedly the cold blast. 



TENSILE, ETC., STRENGTH OF CAST-IRON. 



343 



With the hot blast, unless the materials — i.e., the ore, fuel, 
and flux — are very pure, more of the impurities are fused, and 
combine with the iron ; whereas the iron can be made much 
more rapidly by its use. Indeed, very little cold-blast charcoal 
iron is made at the present time ; i-e., iron where the blast is 
cold, and where charcoal is the only fuel used. 

The rules of the Ordnance Department of the United 
States require that all cast-iron which is used for cannon shall 
have at least a tensile strength of 30000 lbs. per square inch. 

The specimens used for testing the tensile strength of cast- 
iron have generally been made with shoulders : and the smallest 
part has had, in most cases, no length ; and the specimen has 
had, therefore, very little opportunity to stretch. 

Colonel Rosset, of the Arsenal at Turin, made a series of 
experiments upon the influence of the shape of the specimen 
upon the tensile strength. For this purpose he used specimens 
with shoulders ; and, among other tests, he compared the 
strength of the same iron by using specimens the lengths of 
whose smallest parts were respectively 1 metre, 30 millimetres, 
and o millimetres, with the following results : — 



Length of Specimen. 


Tensile Strength, in lbs., per Square Inch. 


1st Cannon. 


2d Cannon. 


3d ■Cannon. 


1 metre . . 

30 millimetres . 

millimetres . 


31291 

3257 1 
33993 


25601 

345 62 
36411 


28019 
3001 1 
3001 1 



It will thus be seen, that, before we can decide upon the 
quality of cast-iron as affected by the tensile strength, it is 
necessary to know the length of that part of the specimen 



344 



APPLIED MECHANICS. 



which has the smallest area. Colonel Rosset's tests of cast- 
iron were almost entirely confined to high-grade irons, suitable 
to use in cannons. 

He deduced, for mean value of the modulus of elasticity of 
the specimens I metre in length, 20419658 lbs. per square inch : 
this, of course, is a modulus only adapted to these high grades, 
and is not applicable to common cast-iron. 

Before proceeding to iron columns, the values given for the 
tensile and compressive strength, and modulus of elasticity, 
of cast-iron, in lbs., per square inch, given by Rankine and 
Weisbach, will be stated ; which values are copied by many 
handbooks : — 







Tensile 


Compressive 


Modulus of 






Strength. 


Strength. 


Elasticity. 






1 13400 


/ 82000 


/ I40OOOOO 


Professor Rankine . 


Various qualities, 


) tC> 


i t0 


to 






( 290OO 


( I45000 


( 22900000 




Average .... 


16500 


I I 2000 


17000000 


Professor Weisbach, 


Average .... 


18500 


- 


14220000 



§219. Cast-Iron Columns. Hodgkinsons Experiments. — 
The high compressive strength of cast-iron would seem to ren- 
der it a very suitable material for all cases of columns or struts. 

Nevertheless, its use for the compression members of bridge 
and roof trusses has now been almost entirely abandoned; and 
wrought-iron has taken its place for these purposes, — partly 
because these struts are very long, and, if of cast-iron, might, 
on account of the bending, bring into play its tensile strength; 
and partly because wrought-iron and steel can be easily riveted, 
joined to other pieces, and repaired, whereas cast-iron cannot 
so easily. 

Cast-iron is still in use very extensively for the columns of 



CAST-IRON COLUMNS. 345 

buildings, as these are generally shorter, and not liable to such 
varying loads, as is the case with bridges. 

Almost the only experiments on the strength of cast-iron 
columns are those made by Eaton Hodgkinson, and recorded 
in the London Philosophical Transactions for 1840. 

These experiments were made by him at the works of Sir 
William Fairbairn, who, as he says, put every means for a full 
investigation into his hands, and expressed the wish that he 
should extend the inquiry to pillars of various kinds, ancient 
as well as modern, and leave no part of the subject in uncer- 
tainty for the want of experiments sufficiently varied and 
extensive. 

The pillars with which the experiments were made were 
mostly of cast-iron, as being the material in most general use 
at that time for that purpose ; but some were of wrought-iron, 
and a few of wood. 

Until the time when Hodgkinson made his experiments, the 
prevailing theory of the strength of columns was that of Euler, 
which has been already explained in § 210. 

According to this theory, a column of any given dimensions 
and material requires a certain weight to bend it, even in the 
slightest degree ; and, with less than this weight, it would not 
be bent at all. 

This weight producing incipient flexure is to be found, as 
we have already seen, by the formulas, 

P = ( — ) EI for one end fixed in direction and the other 



rounded. 



=©■ 



P = [-) EI for both ends rounded. 



/27A 2 

P = ( — J EI for both ends fixed in direction, where / = 

length of column, / — moment of inertia of 
section, E = modulus of elasticity of the ma- 
terial. 



346 APPLIED MECHANICS. 

The load P is, then, according to Euler, the breaking-load, 
unless the column is so short that the breaking-load by direct 
crushing is less than the value of P, in which case the column 
will fail by direct crushing. Starting from this point of view, 
Hodgkinson says, in his report, '" My first object was to supply 
the deficiencies of Euler's theory of the strength of pillars if it 
should appear capable of being rendered practically useful, and, 
if not, to endeavor to adapt the experiments so as to lead to 
useful results." He also says, in regard to Euler's theory, "I 
have many times sought experimentally, with great care, for the 
weight producing incipient flexure according to the theory of 
Euler, but have hitherto been unsuccessful. So far as I can 
see, flexure commences with weights far below those with which 
pillars are usually loaded in practice. It seems to be produced 
by weights much smaller than are sufficient to render it capable 
of being measured." 

"With respect to the conclusions of some writers, that flex- 
ure does not take place with less than about half the breaking- 
weight, this, I conceive, could only mean large and palpable 
flexure ; and it is not improbable that the writers were in some 
degree deceived, from their having generally used specimens 
thicker, compared with their length, than have been usually 
employed in the present effort." 

Another matter to which Hodgkinson devoted considerable 
attention in this investigation, was a comparison of the strength 
of pillars with flat and with rounded ends respectively. By 
flat ends is meant having the ends finished in planes perpen- 
dicular to the axis, and having the resultant of the weight act 
along the axis : the ends are then fixed in direction. By rounded 
ends Hodgkinson actually meant rounded ends ; and he used 
such in his experiments, making them generally hemispherical 
in form. The results have been assumed generally to apply to 
pin ends, or to any irregularity of fixing the ends which docs 
not absolutely fix them in direction. 



CAST-IRON COLUMNS. 347 

As a result of his experiments, he states, that, in all pillars 
whose length is thirty times the diameter or upward, the 
strength of those with flat ends seems to be about three times 
as great as the strength of those with rounded ends ; the mean 
ratio being 3.167. 

In pillars shorter than thirty diameters, the ratio decreased 
when the ratio of the length to the diameter decreased. 

The experiments were all made on circular or hollow cir- 
cular pillars, the lengths varying from j\ diameters to 121 
diameters; a case of the first being one 15 \ inches long and 
0.51 inch diameter, and of the last, 30^ inches long and 1.01 
inch diameter ; the largest diameters used being about two 
inches. 

The empirical formulae given by Hodgkinson for the strength 
of cast-iron columns more than thirty diameters long, are as 
follows ; viz., — 

FOR SOLID CYLINDRICAL COLUMNS. 

(a) With rounded ends, 

F = 33379^-i (1) 

(b) With flat ends, 

P = 98982 J- 6 : (2) 

where d= diameter in inches, / = length in feet, P = breaking- 
weight in pounds. 

FOR HOLLOW CYLINDRICAL COLUMNS. 

(a) With rounded ends, 

P = 2 9 I20 _ ; (3) 



348 APPLIED MECHANICS. 

(b) With flat ends, 

^=993i8 — : (4) 

where D =. outside diameter, d = inside diameter in inches, and 
/ = length in feet. 

For columns less than thirty diameters long, Hodgkinson 
gives the following formula for the breaking-strength : — 

be 

(5) 



where c = force which would crush the pillar without bending 
it, b = breaking-weight that would be obtained by using the 
formula for columns more than thirty diameters long, y = act- 
ual breaking-strength. 

For determining the value of c, we have as the crushing- 
strength of cast-iron, from some experiments of Hodgkinson 
made at the time on cast-iron, the value 109801 lbs. per square 
inch ; so that we should have, 



For solid circular columns, 



4 

For hollow circular columns, 



1 09801- — ; (6) 



7r(Z>-</ 2 ) , v 

c — 109801— i <-. • (7) 



The number 33379 of equation (1) is the mean result from 
eighteen pillars, varying in length from 121 times the diameter 
down to 15 times the diameter; the 98982 of equation (2) is 
similarly obtained from eleven pillars, varying in length from 
yS to 25 times the diameter. 

Mr. James B. Francis has computed and published a series 



TRANSVERSE STRENGTH OF CAST-IRON. 349 

of tables of the strength of cast-iron pillars of the ordinary sizes 
used in practice, the computations being made by means of 
Hodsrkinson's formulae. 

We have also Gordon's formulae, which have been already 
given in § 209, the constants of which are based upon Hodgkin- 
son's experiments. While we have, in the case of wrought-iron 
and wood, the results of experiments made upon full-size col- 
umns, which furnish us more reliable data upon which to base 
our conclusions in designing such columns, we have, in the case 
of cast-iron, no way of obtaining more reliable data than those 
furnished by Hodgkinson's experiments; and hence we can use 
these results only with a large factor of safety. Mr. Francis 
recommends five. 

§220. Transverse Strength of Cast-Iron. — At one time 
cast-iron was very largely used for beams and girders to support 
a transverse load. Its use for this purpose has now been almost 
entirely abandoned, as it has been superseded by wrought-iron. 

A great many experiments have been made on the trans- 
verse strength of cast-iron ; the specimens used in some cases 
being small, and in others large. The records of a great many 
experiments of this kind are to be found in the first four books 
of the list already enumerated in § 218. The details of these 
tests will not be considered here, but an outline will be given 
of some of the main difficulties that arise in applying the results 
and in using the beams. 

That cast-iron is treacherous and liable to hidden flaws ; and 
that it is brittle, is well known by all who use it. It is also a 
fact, that in casting any piece where the thickness varies in 
different parts, the unequal cooling is liable to establish initial 
strains in the metal, and that, therefore, those parts where 
such strains have been established have their breaking-strength 
diminished in proportion to the amount of these strains. These 
defects are enough to exclude cast-iron from use for beams 
when we have a much better material in wrought-iron. 



jy 



APPLIED MECHANICS. 



Another element of uncertainty, which has vexed experi- 
menters, and which has not yet received a complete solution, is 
the fact, that whereas we can, with some approximation to cor- 
rect results, predict the strength of a wrought-iron beam, from 
simply knowing the tensile and compressive strength of the 
iron per square inch, by applying our ordinary theory of beams, 
it is impossible to do so in the case of a cast-iron beam. Thus, 
in the case of a cast-iron which would have a tensile strength 
of 16000 to 18000 lbs. per square inch, and a compressive 
strength of about 80000 lbs. per square inch, the modulus of 
rupture would be variable, but very seldom either of these two 
numbers. 

In Rankine's tables, the modulus of rupture for rectangular 
beams of cast-iron is given on the average as 40000 lbs. per 
square inch, and 'that for openwork beams as 17000. 

What the meaning of this discrepancy was, and how to get 
a uniform modulus of rupture, engaged the attention of many 
experimenters. One explanation offered was, that the neutral 
axis was not at the centre of gravity of the section, but that its 
position was in some way affected by the relation between the 
tensile and the compressive strength of the iron. A moment's 
thought, and a consideration of the assumptions made in the 
common theory of beams, will show the absurdity of any such 
conclusion. As well might we conclude that a loaded beam, 
resting on two supports at the ends, would have its pressure on 
these supports regulated by the strength of the supports, instead 
of by the principle of the lever. 

Besides this, the younger Barlow tried an experiment to 
determine the position of the neutral axis in a rectangular 
cast-iron beam, and found it at the middle of the depth. 

The elder Barlow proposed a very elaborate explanation, 
deducing some much more complicated formulae for the strength 
of beams than the ordinary ones ; but there is not good evidence 
of the truth of this theory. 



WR O UGH T-IRON. 3 5 I 



The most plausible explanation seems to be that offered by 
Rankine; viz., that, besides unequal cooling and the consequent 
establishing of initial strains, the variation is caused by the 
proportion of the hardened, tough skin of the metal. This 
skin theory is borne out, to a great extent, by a series of tests 
made by Edwin Clark, and detailed in D. K. Clark's "Rules 
and Tables," p. 562. 

§221. Wrought-Iron. — Wrought-iron is the product ob- 
tained by removing the carbon from cast-iron. It is produced 
by melting the iron, and passing an oxidizing flame over it. 
When the carbon is burned out, the mass of iron is left in a 
pasty condition, full of holes. It is then taken out, and ham- 
mered or rolled in order to unite it into one mass. Wrought- 
iron is thus, from the commencement of its manufacture, a 
series of welds ; and the perfection or imperfection of these 
welds affects very seriously the quality of the iron. 

The result of this first process is not suitable to use in any 
construction of importance ; but it requires to be re-heated 
and re-rolled a number of times, in order to make it more homo- 
geneous, and to remove flaws from within the iron. 

At best, however, wrought-iron is a series of welds ; and, if 
a piece be broken, the separate layers of which it is composed 
can be seen plainly. It is also subject to the impurities of the 
cast-iron from which it is made. Thus, the presence of sulphur 
makes it red-short, or brittle when hot ; and the presence of 
phosphorus makes it cold-short, or brittle when cold. 

It cannot, like cast-iron, be melted and run into moulds : but 
it can be welded ; that is, two masses of wrought-iron can be 
united by being brought to a proper temperature, and then 
hammered together. 

Wrought-iron is much more capable of bearing a tensile or 
transverse stress than cast-iron : it is tougher, it stretches 
more, and gives more warning before fracture. At one time 
cast-iron was almost the only form in which iron was used in 



352 APPLIED MECHANICS. 

construction ; but now wrought-iron and steel are superseding 
it in by far the majority of <*ases where strength and toughness, 
and the ability to resist varied stresses, are demanded. 

Wrought-iron is also expected to withstand a great many 
trials that would seriously injure cast-iron: thus, two pieces of 
wrought-iron are generally united together by riveting; the 
holes for the rivets have to be punched or drilled, and then 
the rivets have to be hammered ; the entire process tending to 
injure the iron. Wrought-iron has to withstand flanging, and is 
liable to severe shocks when in use ; as, for instance, those that 
occur from the difference of temperature, and the changes of 
temperature in the different parts of a steam-boiler. 

§ 222. Tensile and Compressive Strength of Wrought- 
iron. — As to the experimenters on the tensile strength and 
elasticity of wrought-iron, those who preceded Hodgkinson are 
of little more than historic interest. The following list includes 
a number of the most known experimenters : — 

i°. Eaton Hodgkinson : (a) Report of Commissioners on the Applica- 
tion of Iron to Railway Structures. 
(b) London Philosophical Transactions. 1840. 

2 . William H. Barlow : Barlow's Strength of Materials. 

3 . Sir William Fairbairn : On the Application of Cast and Wrought 
Iron to Building Purposes. 

4 . Franklin Institute Committee : Report of the Committee of the 
Franklin Institute. In the Franklin Institute Journal of 1837. 

5 . L. A. Beardslee, Commander, U.S.N. : Experiments on the Strength 
of Wrought-iron and of Chain Cables. Reyised and enlarged by 
William Kent, M.E., or Executive Document 98, 45th Congress, 
as stated below. 

6°. David Kirkaldy : Experiments on Wrought-iron and Steel. 

7 . Professor Bauschinger : Mittheilungen aus dem Mech.-Tech. Labora- 
torium der K. Pol. Schule in Munchen. 

8°. G. Bouscaren : Report on the Progress of Work on the Cincinnati 
Southern Railway, by Thomas D. Lovett. Nov. 1, 1875. 



TENSILE, ETC., STRENGTH OF WROUGHT-IRON. 353 

Government Testing Machine at Watertown, and Government 
Commission : (a) Executive Document 98, 45th Congress 
U.S.A., 2d session. 

(b) Executive Document 23, 46th Congress U.S.A., 2d session. 

(V) Executive Document 12, 47th Congress U.S.A., 1st session. 

(d) Executive Document 1, 47th Congress U.S.A., 2d session. 

(e) Executive Document 5, 48th Congress U.S.A., 1st session. , 
A. Wohler : (a) Die Festigkeits versuche mit Eisen und Stahl. 

(b) Strength and Determination of the Dimensions of Structures of 
Iron and Steel, by Dr. Phil. Jacob J. Weyrauch. Translated by 
Professor Dubois. 
Alexander Holley : Executive Document 23, 46th Congress 
U.S.A., 2d session. 
12 . Professor R. H. Thurston : Materials of Engineering. 

A few tests were made on the tensile strength of wrought- 
iron by Eaton Hodgkinson : two of these were on the tensile 
strength and elasticity of rods about fifty feet long ; each rod be- 
ing made in three parts, these parts being united by couplings. 

Below is given the table of results of the first of these tests, 
as recorded in the commissioners' report, to which table is added 
here the column of modulus of elasticity : — 



10^ 



11 



Weights 

applied, in 

lbs. 


Exten- 
sion, in 
inches. 


Sets, in 
inches. 

Percep- 
tible 


Weights 
per Square 

Inch of 

Section, in 

lbs. 


Strains, in 

Fractions 

of the 

Length. 


Sets, in Frac- 
tions of 
the Length. 


Ratio of 
Weight 
to Exten- 
sion. 


Modulus 

of 
Elasticity. 


560 


O.O485] 


| 2668 


O.OOO082 


-. 


270544 


32465280 




( 


Percep- 


] 










1 120 


O.IO95J 


tible 

after one 

hour 


\ 5335 
J 


O.OOO185 


- 


239565 


2829780O 


l68o 


O.1675 


0.0015 


8003 


O.OOO284 


O.OOOOO25 


234889 


28186680 


224O 


O.2240 


0.0020 


10670 


O.OOO379 


O.OOOOO34 


234201 


2810412O 


2S0O 


O.2S05 


0.0027 


13338 


O.OOO475 


O.OOOQO42 


233791 


28054920 


3360 


3370 


0.0030 


16005 


O.OOO57I 


O.OOOOO47 


233518 


27017160 


3920 


O.3930 O.OO4O 


18673 


0.000666 


O.OOOO068 


233616 


28033920 


44S0 


O.4520 O.OO75 


21340 


O.OOO766 


O.OOOOI27 


232138 


27856560 


504O 


O.5155 | O.OI95 


24008 


O.OOO874 


O.OOOO330 


22S975 


27477OOO 



354 



APPLIED MECHANICS. 



Weights 

applied, in 

lbs. 


Exten- 
sion, in 
inches. 


Sets, in 
inches. 


Weights 
per Square 

Inch of 

Section, in 

lbs. 


Strains, in 

Fractions 

of the 

Length. 


Sets, in Frac- 
tions of 
the Length. 


Ratio of 
Weight 
to Exten- 
sion. 


Modulus 

of 
Elasticity. 


5600 


O.5980 


O.049O 


26676 


O.OOIOI4 


O.OOO0830 


219317 


26318040 


6160 


O.760O 


0.1545 


29343 


O.OOI288 


O.OOO2619 


189825 


227790OO 


6720 


I.3IOO 


— 


320I I 


0.002228 


— 


— 


— 


In 10 min. 


I.3780 


O.6670 


- 


O.OO2356 


O.OO 1 1305 


I 13228 


13587360 


In 12 min. 


I.390O 


— 


— 


— 


— 


— 


— 


7280 


2.53!° 


I.8125 


34678 


O.OO4290 


O.OO30720 


67363 


8083560 


7840 


5.4060 


5.0000 


37346 


O.OO9163 


O.O084746 


33966 


4075920 


Repeated 


5-8750 


5.0625 


- 


O.OO9957 


O.O085805 


- 


- 


8400 


6.219O 


5-3750 


4OOI3 


O.OIOI66 


O.OO91083 


32798 


3935760 


In 1 hour 


6.937O 


- 


- 


O.OII758 


- 


- 


- 


In 2 hours 


7.0000 


- 


- 


0.0 1 1866 


- 


- 


- 


In 3 hours 


7.O47O 


- 


- 


O.OII775 


- 


- 


- 


In 4 hours 


7.O5OO 


- 


— 


O.O I I95O 


- 


- 


- 


In 5 hours 


7.062O 


- 


- 


O.0 1 1966 


- 


- 


- 


In 6 hours 


7.064O 


- 


- 


O.OII975 


- 


- 


- 


In 7 hours 


7.O94O 


- 


- 


O.OI2025 


- 


- 


- 


In 10 hours 


7.O94O 


- 


- 


O.OI2025 


- 


— 


- 


Left over 


1 














night ; 


\ 7.OOOO 














next morn- 














ing 

8960 


J 
IO.S620 


9.7500 


42681 


O.OI79OO 


O.O165250 


19870 


2384400 


In 5 min. 


II.5OOO 


- 


- 


O.OI9492 


- 


- 


- 


In 10 min. 


II.7I9O 


- 


- 


O.OI9858 


- 


- 


- 


In 1 hour 


- 


10.875 


- 


- 


OO184333 


- 


- 


In 46 hours 


II.937O 


11,0000 


- 


O.O20233 


O.0 1 864 1 7 


- 


- 


9520 


I2.687O 


11.6880 


45348 


O.O2150O 


O.OI98083 


17577 


2109240 


In 1 hour 


I2.8IOO 


- 


- 


O.O21708 


- 


- 


- 


In 2 hours 


I2.8150 


- 


- 


O.O21716 


- 


- 


- 


In 19 hours 


I2.815O 


11.8150 


- 


O.O21716 


O.02O0250 


- 


- 


10080 


I4.625O 


- 


48016 


O.O24792 


- 


- 


- 


In 10 min. 


- 


I3-4370 


- 


- 


O.O227750 


16139 


I936680 


In 1 hour 


I4.844O 


- 


- 


0,025158 


- 


- 


- 


In 1 1 hours 


I4.89IO 


- 


- 


O.O25242 


- 


- 


- 


Next morn- 
ing 


[14.8750 


- 


- 


- 




- 


- 


10640 


_ 


_ 


_ 


— 


— 


— 


— 


In 10 min. 


2O.625O 


19-375° 


50684 


O.O34958 


O.O328417 


12082 


1449840 


Repeated 


20.7780 


- 


- 


O.O35217 


- 


- 


- 


In 7 hours 


20.7810 


- 


- 


O.O35225 


- 


- 


- 


In 12 hours 


20.7810 




- 


O.O35225 


- 


- 


- 


Next morn- 


) 














ing 


?■ 20-7 5OO 














1 1 200 






53351 











TENSILE, ETC., STRENGTH OF WROUGHT-IRON 355 

With this last weight the rod broke at one of the weldings, 
where there was a slight defect ; perhaps a rather smaller weight 
would have broken it. 

In connection with this table of results, the following ob- 
servations will be made: — 

i°. The breaking-strength per square inch in this case was 
53351 lbs. per square inch. 

2 . Permanent set began with very small loads. 

3 . The limit of elasticity was about 29000 lbs. per square 
inch, this being about the load where permanent set began to 
increase rapidly. 

4 . The modulus of elasticity of the rod in this case was 
about 28000000 lbs. per square inch, this being the ratio of the 
stress to the strain under loads less than the limit of elasticity. 

5 . This experiment shows, that, under moderate loads, the 
modulus of elasticity, and hence the ratio of the stress to the 
strain, remains much more nearly constant with wrought than 
with cast iron. 

6°. The other experiment, with a rod 50 feet long, gave as 
modulus of elasticity 27691200 lbs. per square inch. 

The remaining experiments of Hodgkinson on wrought- 
iron may be found in the report of the commissioners already 
referred to. 

Barlow f s experiments on wrought-iron will not be detailed 
here, except only to say that he tried seven experiments to 
determine the modulus of elasticity of wrought-iron, and that 
he obtained, for mean extension per English ton per square inch, 
'maximum, 0.0001082 ; minimum, 0.0000841 ; mean, 0.0000956 
of the length. These correspond to moduli of elasticity re- 
spectively equal to 20702400, 26634900, and 23430900 lbs. per 
square inch. 

The results obtained by experimenters before Hodgkin- 
son's time are very discordant and very uncertain, many of 
them attributing to wrought-iron a strength far greater than 



356 APPLIED MECHANICS. 

can be attained at the present time. There is no satisfactory 
evidence, however, to show that the wrought-iron of that time 
was any better than (if as good as) that made at the present 
time ; and it is more probable either that there were errors in 
making the tests, or else that the supposed iron was really steel, 
and possibly brittle. The paucity of the records, and the im- 
possibility of obtaining the details of the tests, render any 
search for the reason in any special cases futile, and throw 
doubt upon the greater part of these tests. 

Among the later English experimenters, we have Sir William 
Fairbairn. An account of his tests on tensile strength will be 
found in the book already referred to. 

One of the most prominent English experimenters, and one 
who has done a great deal towards rendering our knowledge of 
this subject more accurate, is David Kirkaldy. His results up 
to 1866 are detailed in his book entitled, "Experiments on 
Wrought-iron and Steel," published in that year. 

In the early part of his book will be found a summary of 
what had been done in this line by the early experimenters. 

Kirkaldy tested a large number of English irons ; and a 
summary of his results will be given here, together with his 
sixty-six concluding observations. 



I 



TENSILE, ETC., STRENGTH OF WROUGHT-IRON 357 









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360 



APPLIED MECHANICS. 



SUMMARY OF KIRKALDY'S RESULTS. — IRON PLATES. 



Names of Makers or Works. 



Farnley ...... 

Farnley 

Lowmoor 

Farnley 

Bowling 

<3> Govan O . . . 
Bradley & Co. S. C. 
Bradley & Co., L. F. . 
Bradley & Co. ... 
Malinslee Best . . 
Consett Best Best . . 
Consett Best Best . . 
Consett Best Best . . 
Thorneycroft Best Best 
Snedshill ^ Best . . 
Wells Best ^ Best . 
Glasgow Best Boiler . 



o 

si 
3° 



L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 



Breaking-Weight per 
Square Inch of Area. 



Original, 
lbs. 



58437 
55033 
58487 
54098 
52000 

50515 
56005 
4622I 

52235 
46441 

54644 

49399 
55831 
50550 
56996 
5 I2 5 J 
55708 

49425 
52572 
50627 

51245 
46712 

53559 
45677 
49120 

46755 
54847 
45585 
52362 
43036 
45997 
493" 
53849 
48848 



Fractured, 
lbs. 



83I 12 
68961 

70538 
59698 
64746 

57383 
68763 

53293 
61716 
5OOO9 
66728 
54020 
67406 
55206 
66858 
5607O 
65652 
54002 
6213I 
55746 
59183 
52050 
62346 
48358 
55472 
5OOOO 
62747 
47712 
61581 

45300 
51140 

54842 
60522 
52252 



Contrac- 
tion of 
Area at 
Fracture, 
per cent. 



29.6 
20.I 
20.0 
II.4 
19.7 
I2.I 
17.8 
13.2 

15-3 

6.9 

18. 1 

8-5 
17.2 

9.0 
15.0 

8.6 
14.9 

8.1 
15.4 

9-3 
i3-i 
10.2 
14.4 

5.6 

"•3 

6-3 
12.5 

4.6 
15.0 

5-3 
10.4 
10.5 
11.0 

6.6 



Extreme 
Elonga- 
tion, 
per cent. 



17.O 

"•3 
10.9 

5-9 
13.2 

9-3 
14.1 

7.6 
11.6 

5-9 
11.6 

6.5 
12.5 

5-5 
13.0 

5-9 

10.7 

5-i 
8.6 

5.8 
8.9 
6.4 

"•5 

4.0 
8.0 
5.2 
11. 2 
4.6 
9.6 
2.8 
6.7 
7.0 

9-3 

4.6 



KIRKALDY'S SIXTY-SIX CONCLUSIONS. 



36l 



SUMMARY OF KIRKALDY'S RESULTS. — IRON PLATES. — Concluded. 



Names of Makers or Works. 



c o 

,3 



Breaking- Weight per 
Square Inch of Area. 



Original, 
lbs. 



Fractured, 
lbs. 



Contrac- 
tion of 
Area at 
Fracture, 
per cent. 



Extreme 
Elonga- 
tion, 
per cent. 



Glasgow Best Best . . . 
Wells Best yjjf Best . . 
Maker's stamp uncertain . 

K. B. M. 

Lloyds, Foster, & Co., Best 
Glasgow Best Best . . . 
Glasgow Ship .... 
Mossend Best Best . 

Govan Best 

Glasgow Best Scrap . . 



53399 
41791 
474io 
46630 

47598 
40682 
46404 
44764 
44967 

4473 2 
45626 

41340 
47773 
44355 
43433 
4H5 6 
43942 
39544 
50844 



59557 
43614 

51521 

48348 
53182 
43426 
51896 
47891 
49162 

48344 
48208 
42430 
49816 

45343 
46038 
43622 
45886 
40624 
58412 



10.6 

3-7 
67 
4.9 
9.8 

5-o 
10.2 
6.4 
8.7 
6.9 

5-3 
2.2 
4.8 
3-4 
5-7 
4.9 
6.0 
2.6 
13.0 



9.0 
2.6 
4.0 
3-4 
5-9 
2-5 
6.1 

4-3 
5-3 
4.6 

4-3 
2.4 

3-6 
2.1 

3-3 
2.9 

3-4 

1.4 

10.5 



Fracture 1-12. — Light gray of various shades, close and very fine. 
Fracture 13-30. — Dullish gray of various shades, close and generally fine. 
Fracture 31-44. — Dull gray, generally rather coarse and irregular. 
Fracture 45—52. — Irregular, generally coarse and open. 
Fracture 53. — Close and fine. 



KIRKALDY'S SIXTY-SIX CONCLUSIONS. 

i°. The breaking-strain does not indicate the quality, as hitherto 
assumed. 

2 . A high breaking-strain may be due to the iron being of superior 
quality, dense, fine, and moderately soft, or simply to its being very 
hard and unyielding. 



362 APPLIED MECHANICS. 

3 . A low breaking-strain may be due to looseness and coarseness 
in the texture, or to extreme softness, although very close and fine in 
quality. 

4 . The contraction of area at fracture, previously overlooked, 
forms an essential element in estimating the quality of specimens. 

5 . The respective merits of various specimens can be correctly 
ascertained by comparing the breaking-strain jointly with the contrac- 
tion of area. 

6°. Inferior qualities show a much greater variation in the break- 
ing-strain than superior. 

7 . Greater differences exist between small and large bars in coarse 
than in fine varieties. 

8°. The prevailing opinion of a rough bar being stronger than a 
turned one, is erroneous. 

9 . Rolled bars are slightly hardened by being forged down. 

io°. The breaking-strain, and contraction of area, of iron plates are 
greater in the direction in which they are rolled than in a transverse 
direction. 

n°. A very slight difference exists between specimens from the 
centre and specimens from the outside of crank-shafts. 

12 . The breaking-strain, and contraction of area, are greater in those 
specimens cut lengthways out of crank-shafts than in those cut cross-ways. 

1 3 . The breaking-strain of steel, when taken alone, gives no clew 
to the real qualities of various kinds of that material. 

1 4 . The contraction of area at fracture of specimens of steel must 
be ascertained, as well as in those of iron. 

1 5 . The breaking-strain, jointly with the contraction of area, affords 
the means of comparing the peculiarities in various lots of specimens. 

1 6°. Some descriptions of steel are found to be very hard, and con- 
sequently suitable for some purposes ; whilst others are extremely soft, 
and equally suitable for other uses. 

1 7 . The breaking-strain, and contraction of area, of puddled steel 
plates, as in iron plates, are greater in the direction in which they are 
rolled ; whereas in cast-steel they are less. 

1 8°. Iron, when fractured suddenly, presents invariably a crystalline 
appearance ; when fractured slowly, its appearance is invariably fibrous. 



KIRKALDY'S SIXTY-SIX CONCLUSIONS. 363 

1 9 . The appearance may be changed from fibrous to crystalline by 
merely altering the shape of the specimen so as to render it more liable 
to snap. 

20 . The appearance may be changed by varying the treatment so 
as to render the iron harder and more liable to snap. 

21 . The appearance may be changed by applying the strain so 
suddenly as to render the specimen more liable to snap from having less 
time to stretch. 

22 . Iron is less liable to snap the more it is worked and rolled. 

23 . The "skin," or outer part of the iron, is somewhat harder than 
the inner part, as shown by appearance of fracture in rough and turned 
bars. 

2 4 . The mixed character of the scrap-iron used in large forgings 
is proved by the singularly varied appearance of the fractures of speci- 
mens cut out of crank-shafts. 

25 . The texture of various kinds of wrought-iron is beautifully 
developed by immersion in dilute hydrochloric acid, which, acting on 
the surrounding impurities, exposes the metallic portion alone for exam- 
ination. 

2 6°. In the fibrous fractures the threads are drawn out, and are 
viewed externally; whilst in the crystalline fractures the threads are 
snapped across in clusters, and are viewed internally or sectionally. In 
the latter cases, the fracture of the specimen is always at right angles 
to the length ; in the former, it is more or less irregular. 

2 7 . Steel invariably presents, when fractured slowly, a silky, fibrous 
appearance ; when fractured suddenly, the appearance is invariably 
granular, in which case also the fracture is always at right angles to the 
length ; when the fracture is fibrous, the angle diverges always more or 
less from 90 . 

2 8°. The granular appearance presented by steel suddenly fractured 
is nearly free from lustre, and unlike the brilliant crystalline appearance 
of iron suddenly fractured : the two combined in the same specimen 
are shown in iron bolts partly converted into steel. 

29 . Steel which previously broke with a silky, fibrous appearance is 
changed into granular by being hardened. 

30 . The little additional time required in testing those specimens 



364 APPLIED MECHANICS. 

whose rate of elongation was noted, had no injurious effect in lessening 
the amount of breaking-strain, as imagined by some. 

31 . The rate of elongation varies not only extremely in different 
qualities, but also to a considerable extent in specimens of the same 
brand. 

3 2 . The specimens were generally found to stretch equally through- 
out their length until close upon rupture, when they more or less sud- 
denly draw out ; usually at one part only, sometimes at two, and in a few 
exceptional cases at three, different places. 

33 . The ratio of ultimate elongation may be greater in short than 
in long bars in some descriptions of iron, whilst in others the ratio is not 
affected by difference in the length. 

34 . The lateral dimensions of specimens form an important ele- 
ment in comparing either the rate of or the ultimate elongations, — a 
circumstance which has been hitherto overlooked. 

35 . Steel is reduced in strength by being hardened in water, while 
the strength is vastly increased by being hardened in oil. 

36 . The higher steel is heated (without, of course, running the risk 
of being burned), the greater is the increase of strength by being 
plunged into oil. 

37 . In a highly converted or hard steel the increase in strength 
and in hardness is greater than in a less converted or soft steel. 

38 . Heated steel, by being plunged into oil instead of water, is not 
only considerably hardened, but toughened, by the treatment. 

39 . Steel plates hardened in oil and joined together with rivets are 
fully equal in strength to an unjointed soft plate, or, the loss of strength 
by riveting is more than counterbalanced by the increase in strength by 
hardening in oil. 

40 . Steel rivets fully larger in diameter than those used in riveting 
iron plates of the same thickness, being found to be greatly too small for 
riveting steel plates, the probability is suggested that the proper propor- 
tion for iron rivets is not, as generally assumed, a diameter equal to the 
thickness of the two plates to be joined. 

41 . The shearing-strain of steel rivets is found to be about a fourth 
less than the tensile strain. 

4 2 . Iron bolts case-hardened bore a less breaking-strain than when 



KIRKALDY'S SIXTY-SIX CONCLUSIONS. 365 

wholly iron, owing to the superior tenacity of the small proportion of 
steel being more than counterbalanced by the greater ductility of the 
remaining portion of iron. 

43 . Iron highly heated, and suddenly cooled in water, is hardened, 
and the breaking-strain, when gradually applied, increased ; but at the 
same time it is rendered more liable to snap. 

44 . Iron, like steel, is softened, and the breaking-strain reduced, 
by being heated, and allowed to cool slowly. 

45 . Iron subjected to the cold-rolling process has its breaking- 
strain greatly increased by being made extremely hard, and not by being 
"consolidated," as previously supposed. 

46 . Specimens cut out of crank-shaft are improved by additional 
hammering. 

47 . The galvanizing or tinning of iron plates produces no sensible 
effects on plates of the thickness experimented on. The results, how- 
ever, may be different should the plates be extremely thin. 

48 . The breaking-strain is materially affected by the shape of the 
specimen. Thus, the amount borne was much less when the diameter 
was uniform for some inches of the length than when confined to a 
small portion, — a peculiarity previously unascertained and not even 
suspected. 

49 . It is necessary to know correctly the exact conditions under 
which any tests are made, before we can equitably compare results 
obtained from different quarters. 

50 . The startling discrepancy between experiments made at the 
Royal Arsenal and by the writer is due to the difference in the shape 
of the respective specimens, and not to the difference in the two testing- 
machines. 

51 . In screwed bolts, the breaking-strain is found to be greater 
when old dies are used in their formation than when the dies are new, 
owing to the iron becoming harder by the greater pressure required in 
forming the screw-thread when the dies are old and blunt than when 
new and sharp. 

5 2 . The strength of screw-bolts is found to be in proportion to 
their relative areas ; there being only a slight difference in favor of the 
smaller compared with the larger sizes, instead of the very material 
difference previously imagined. 



366 APPLIED MECHANICS. 

53 . Screwed bolts are not necessarily injured, although strained 
nearly to their breaking-point. 

54 . A great variation exists in the strength of iron bars which have 
been cut and welded : whilst some bear almost as much as the -uncut 
bar, the strength of others is reduced fully a third. 

55 . The welding of steel bars, owing to their being so easily burned 
by slightly over-heating, is a difficult and uncertain operation. 

5 6°. Iron is injured by being brought to a white or welding heat 
if not at the same time hammered or rolled. 

57°. The breaking-strain is considerably less when the strain is 
applied suddenly, instead of gradually, though some have imagined that 
the reverse is the case. 

5 8°. The contraction of area is also less when the strain is suddenly 
applied. 

59 . The breaking-strain is reduced when the iron is frozen. With 
the strain gradually applied, the difference between a frozen and un- 
frozen bolt is lessened as the iron is warmed by the drawing-out of the , 
specimen. 

6o°. The amount of heat developed is considerable when the 
specimen is suddenly stretched, as shown in the formation of vapor 
from the melting of the layer of ice on one of the specimens, and also 
by the surface of others assuming tints of various shades of blue and 
orange, not only in steel, but also, although in a less marked degree, in 
iron. 

6i°. The specific gravity is found generally to indicate pretty cor- 
rectly the quality of specimens. 

62 . The density of iron is decreased by the process of wire-drawing, 
and by the similar process of cold-rolling, instead of increased, as pre- 
viously imagined. 

63 . The density in some descriptions of iron is also decreased by 
additional hot-rolling in the ordinary way : in others the density is very 
slightly increased. 

64 . The density of iron is decreased by being drawn out under a 
tensile strain, instead of increased, as believed by some. 

65 . The most highly converted steel does not, as some may sup- 
pose, possess the greatest density. 



TESTS OF COMMANDER BEARDSLEE. 367 

66°. In cast-steel the density is much greater than in puddled steel, 
which is even less than in some of the superior descriptions of wrought- 
iron. 



TESTS OF COMMANDER BEARDSLEE. 

One of the most valuable sets of tests of wrought-iron 
is that obtained by committees D, H, and M of the Board 
appointed by the United-States Government to test iron and 
steel ; the special duties of these committees being to test such 
iron as would be used in chain-cable, and the chain-cable itself. 
The chairman of these three committees, which were consoli- 
dated into one, was Commander L. A. Beardslee of the United- 
States Navy. The full account of the tests is to be found in 
Executive Document 98, 45 th Congress, second session ; and 
an abridged account of them was published by William Kent, 
as has been already mentioned. 

The samples of bar-iron tested were round, and varied from 
one inch to four inches in diameter. The mills from which the 
samples- tested came were as follows : — 

Bentoni Pennsylvania. 

Burden and Sons New York. 

Burgess Ohio. 

Catasauqua Pennsylvania. 

New-Jersey Iron and Steel Company . . . New Jersey. 

Niles Iron Company Ohio. 

Pembroke Massachusetts. 

Pencoyd Pennsylvania. 

Phcenix Pennsylvania. 

Sligo Pennsylvania. 

Tamaqua Pennsylvania. 

Tredegar Virginia. 

Trego and Thompson Maryland. 

Wyeth Brothers Maryland. 



368 APPLIED MECHANICS. 

Certain conclusions which they reached refer to all kinds 
of wrought-iron, and will be given here before giving a table of 
the results of the tests. 

i°. Kirkaldy considers the breaking-strength per square 
inch of fractured area as the main criterion by which to deter- 
mine the merits of a piece of iron or steel. Commander 
Beardslee, on the other hand, thinks that a better criterion is 
what he calls the "tensile limit;" i.e., the maximum load the 
piece sustains divided by the area of the smallest section when 
that load is on, i.e., just before the load ceases to increase in 
the testing-machine. 

2°. Kirkaldy had already called attention to the fact that 
the tensile strength of a specimen is very much affected by its 
shape, and that, in a specimen where the shape is such that 
the length of that part which has the smallest cross-section is 
practically zero (as is the case when a groove is cut around 
the specimen), the breaking-strength is greater than it is when 
this portion is long ; the excess being in some cases as much 
as 33 per cent. 

Commander Beardslee undertook, by actually testing speci- 
mens whose smallest areas varied in length, to determine what 
must be the least length of that part of the specimen whose 
cross-section area is smallest, in order that the tensile strength 
may not be greater than with a long specimen. The conclusion 
reached was, that no test-piece should be less than one-half inch 
in diameter, and that the length should never be less than four 
diameters ; while a length of five or six diameters is necessary 
with soft and ductile metal in order to insure correct results. 
The following results of testing steel are given in Mr. Kent's 
book, as confirming the same rule in the case of steel. The 
tests were make upon Bessemer steel by Col. Wilmot at the 
Woolwich arsenal. 



TESTS OF COMMANDER BEARDSLEE. 



369 



By groove form 
By cylinder . . 



Tensile Strength. 



Highest 

Lowest 

Average 

Highest 

Lowest 

Average 



Pounds per 
Square Inch. 



162974 
136490 

J53677 
123165 

103255 
1 14460 



3 . Commander Beardslee also noticed that rods of certain 
diameters of the same kind of iron bore less in proportion than 
rods of other diameters ; and, after searching carefully for the 
reason, he found it to lie in the proportion between the diam- 
eter of the rod and the size of the pile from which it is 
rolled. The following examples are given : — 

i|-fn. diameter, 6.62% of pile, 56543 lbs. per sq. in. tensile strength. 



*l 


a 


8.18% 


tt 


56478 " 


tt 


tt 


it 


If 


it 


9.90% 


tt 


54277 " 


it 


tt 


it 


l£ 


it 


II. 78% 


it 


53550 " 


tt 


a 


tt 


If 


it 


7.68% 


(t 


56344 " 


tt 


tt 


ft 


If 


tt 


8.90% 


11 


550I8 « 


tt 


tt 


tt 


If 


ti 


10.22% 


ti 


54034 " 


tt 


tt 


tt 


2 


tt 


H.63% 


it 


51848 " 


tt 


tt 


a 



He therefore claims, that, in any set of tests of round iron, 
it is necessary to give the diameter of the rod tested, and not 
merely the breaking-strength per square inch. 

4 . He gives evidence to show, that if a bar is under-heated, 
it will have an unduly high tenacity and elastic limit; and that 
if it is over-heated, the reverse will be the case. 



370 



APPLIED MECHANICS. 



5°. The discovery was made independently by Commander 
Beardslee and Professor Thurston, that wrgught-iron, after 
having been subjected to its ultimate tensile strength without 
breaking it, would, if relieved of its load and allowed to rest, 
have its breaking-strength and its limit of elasticity increased. 

He tried a considerable number of experiments, studying 
the action of this law under different periods of rest, from I 
minute to 3 days and upwards ; and a great deal of valuable 
information is given in the tables. of the report. 

The most characteristic table is the following : — 

EFFECT OF EIGHTEEN HOURS' REST ON IRONS OF WIDELY DIFFER- 
ENT CHARACTERS. 





Ultimate 


Strength 






per Square Inch. 










Remarks. 




First 


Second 




Strain. 


Strain. 




Boiler iron . . . 


48600 


5 6 5°° 


Not broken. 


tt it 


49800 


57000 


Broken -\ 


tt it 


49800 


58000 


Broken I Average gain, 


a a 


48100 


54400 


Broken | 15-8%. 


. t( te 


48150 


5555° 


Broken ) 


Contract chain iron, 


50200 


54000 


Broken "") 

Not broken 1 Average 


a it it 


5° 2 5° 


53200 


it it it 


50700 


553oo 


Not broken Y ' gain, 
Not broken 1 6.4%. 


it it a 


49600 


52900 


it a tt 


51200 


52800 


Not broken J 


Iron K . . . . 


58800 


64500 


Broken ^ 

Broken I Avera S e & m > 

Broken J 9A%- 


a it 
a tt 


59000 
56400 


65800 
60600 



TESTS OF COMMANDER BEARDSLEE. 



371 



His experiments show that the increase is in irons of a 
fibrous and ductile nature, rather than in brittle and steely ones : 
hence the latter class would be but little benefited by the action 
of this law. 

The following table of results is given in the report ; showing 
how the strength per square inch varies with the diameter of 
the piece, in consequence of the amount of reduction in the 
rolls. 



in 






Elastic 


u? 






Elastic 


u? 






Elastic 


M 


B 


Strength 


Limit 1 


O 


G 


Strength 


Limit 





C 


Strength 


Limit 


_C 


8 


per 




G 


g 


per 




.5 


8 


per 




| 


V 

S 


Square 

Inch of 

Original 

Area. 


per 
Square 
Inch of 
Original 


V 

s 




I 

ft 


Square 

Inch of 

Original 

Area. 


per 
Square 
Inch of 
Original 


u 

H 
« 


O 

I 


Square 

Inch of 

Original 

Area. 


per 
Square 
Inch of 
Original 


P 






Area. 


P 






Area. 


P 






Area. 


T 


F 


59885 






D 


52900 




T l 
l 4 





57977 


31996 


a. 

8 


F 


54090 


40980 




F 


52819 


32267 ; 


T 1 
I4 


p 


55782 


35596 


i 


C 


62700 






F 


51400 


34600 ; 


T 1 
*4 


Px 


56334 


33921 


1 
i 


C 


59000 




T 1 
*8 


K 


60458 


37344 


J 4 


N 


56478 


33 2 5I 


* 


C 


57700 




I* 


D 


59582, 


33597 


*4 


Fxl 


55253 


34784 


* 


C 


55400 




1 8 


C 


57470 


31900 ] 


I4 


D 


55550 


28166 


i 


F 


52275 


39126 


1 8 


Fxl 


5 6 434 


34682 


1-4- 


E 


53893 


32712 


1 


F 


55450 




T 1 
*8 


P 


57498 


413" i 


li 


Fx2 


55132 


38603 


5. 

8 


F 


52050 




l Z 


N 


56143 


32267 j 


Ii 


Fx3 


53247 


32520 ; 


f 


F 


57660 




r 1 
*8 


Fx2 


55927 


37250 


Ii 


A 


53897 


27643 


3 

4 


F 


5^46 


35933 


H 


E 


53097 


33549 


x 4 


M 


53752 


- 


1 
8 


F 


50630 


33931 


ii 


Fx3 


54644 


34695 


Ii 


M 


54090 


j 


I 


K 


61727 


1 


it 


D 


54687 


28166 ! 


T 1 
I4 


F 


52970 


32075 | 


I 


D 


61115 


33486 


r ¥ 


A 


539O0 


26787 


il 
x 4 


F 


52729 


39608 i 


I 





57363 


37415 


l T 


F 


53850 


33457 


T l 
*4 


M 


53022 


; 


I 


Fxl 


55768 


34729 ! 


Ii 





53035 


32410 ! 


Ii 


F 


52620 


33220 j 


I 


P 


57807 


39230 


in 


F 


50149 


35493 


T 1 
1 4 





50040 


30730 


I 


A 


54690 


34881 


Is 


F 


52267 


32019 


i-h 


P 


545i8 


35898. 


I 


Fx2 


56790 


36885 


Ii 


K 


59461 


36501 


if 


M 


58926 


37548 


I 


Fx3 


53915 


36336 


T i 

X 4 


P 


56876 


36868 j 


if 


M 


57649 


38578 • 


I 


F 


51921 


31300 


T I 
l 4 


C 


57897 


32469 ; 


*8 


D 


58021 


32152 : 



372 



APPLIED MECHANICS. 



V 

•s 

.s 
# g 

4) 
4J 

B 

a 

s 


a 
o 

V. 

o 

1 


Strength 

per 
Square 
Inch of 

Original 
Area. 


Elastic 
Limit 
per 
Square 
Inch of 
Original 
Area. 


V 

.£ 

w 

E 

s 


C 
O 

O 

u 

i 
55 


Strength 

per 
Square 
Inch of 

Original 
Area. 


Elastic 
Limit 
per 
Square 
Inch of 
Original 
Area. 


in 

u 

y 

a 

_c 

l-T 
u 

i 

s 


C 

6 


Strength 

per 
Square 
Inch of 

Original 
Area. 


Elastic 
Limit 
per 
Square 
Inch of 
Original 
Area. 


If 


K 


55790 


31034 


I* 


M 


54095 


35544 


T 5 - 

1 8 


Fx2 


53438 


35870 


li- 


C 


54949 


31030 


tJ, 


E 


54544 


33027 


T 5 - 


H 


523H 


29364 


lt 


M 


54373 


35820 


,1 


P 


52868 


29636 


A 8 


E 


51946 


27695 


If 


N 


54277 


33622 


tl 
*■■£ 


M 


53512 


- 


x 8 





52401 


34012 


r| 


Fxl 


52968 


33275 


i* 


M 


52941 




1 8 


F 


52163 


33907 


i| 


Fx3 


52733 


34606 


ii 


Fx3 


52819 


34840 


, ft. 


G 


51205 


33318 


if 


E 


52254 


25930 


1* 


Fxl 


53491 


34307 


I- 


F 


50529 


35390 


1 8 


A 


53557 


33650 


T 1 
12 


M 


52736 


34901 


1 8 


F 


50970 


33625 


If 


P 


52556 


30802 


1* 


N 


53555 


34690 


I» 


C 


49030 


31099 


i! 


F; 


52537 


34469 


it 


C 


52700 


3588o 


iH 


K 


56595 


383IO 


it 


F 


52339 


39103 




H 


52462 


29992 


T 1 1 

1 It) 


B 


54l8l 




it 


M 


53°!6 


35379 




D 


52155 


27708 


1 1 6 


J 


54II4 




if 


Fx2 


51487 


359i 1 


I* 


A 


51884 


28794 


»tt 


B 


52895 


33M5 


if 


F 


51296 


31992 


T 1 


-F 


51994 


32054 


I« 


E 


52120 


35549 


if 





5°594 


34940 


n 





50919 


32312 


lii 


G 


57789 


34i6o 


*-h 


P 


53345 


- 


li 


F 


5H56 


34591 


1 lb 


C 


49821 


33i84 


i-k 


E 


53944 


32542 


,1 

1 1 


Fx2 


51481 


34917 


If 


K 


57874 




*A 


G 


53238 


32534 


A 


J 


5io47 


- 


If 


Px 


54212 


•33908 


i'A 


B 


52287 


3241 1 


i* 


M 


49292 


32597 


x 4 


C 


54410 


31354 


i^ 


C 


51756 


32655 


x 8 


N 


56344 


35889 


*4 


P 


52844 


33842 


*ft 


J 


50400 


- 


If 


K 


57132 


35026 


x 4 


Fxl 


53846 


36573 




M 


57052 


38417 


1 8 


M 


57402 


35701 


*4 


H 


53800 


27856 


4 


K 


57317 


33412 


T ft 

1 8 


P 


55634 


33522 


*4 


N 


550I8 


34283 


ii 


D 


56505 


32496 


If 


C 


56227 


33207 


*4 


D 


53472 


31892 


It 


M 


55466 


3478o 


T 5 - 

1 8 


Px 


54689 


33427 


x 4 , 


J 


53264 


- 


17 


M 


55i3i 


33771 


j ft. 


A 


54334 


32163 


If 


D 


52699 


27817 


ft 


P 


54159 


33 J 40 


T 5 - 

1 8 


D 


53695 


30087 


If 


Fx3 


53154 


35323 


ii 


M 


54540 


- 


t 5 - 

1 8 


Fx3 


53339 


33540 


i4 

*4 


E 


51606 


26541 


■ii 


C 


55404 


3477o 


1 8 


Fxl 


53537 


34335 


*4 


A 


51509 


29404 


*2 


E 


55415 


32869 


I» 


D 


53614 


30664 


f 4 


F 


50690 


32229 


1 -l 


M 


54816 


347i6 


T 5 - 


J 


52748 


- 


1 4 


G 


50395 


36254 


n 


Px 


54354 


34617 


If 


E 


52675 


33745 


x 4 


C 


50312 


30852 



TESTS OF COMMANDER BEARDSLEE. 



373 



J2 

E 



C 

o 

o 
g 

rt 


Strength 

per 
Square 
Inch of 

Original 
Area. 


Elastic 
Limit 
per 
Square 
Inch of 
Original 
Area. 


V 

' I 

3 


S 

8 

I 

(4 


Strength 

per 
Square 
Inch of 

Original 
Area. 


Elastic 
Limit 
per 
Square 
Inch of 
Original 
Area. 


1 

.2 

# C ' 

u 
g 

rt 

5 


c 
8 

O 

U 

I 


Strength 

per 
Square 
Inch of 

Original 
Area. 


Elastic 
Limit 

per 

Square 

Inch of 

Original 

Area. 


if 


F 


50547 


35954 


HI 


F 


49355 


32855 


2 





48249 


3I4I3 


if 


Fx2 


523H 


353 2 o 


if 


F 


48670 


23250 


2 


D 


49146 


33068 


if 


E 


49816 


31214 


T 7 





47478 


30842 


2 


D 


46I5I 


36050 


if 


F 


49738 


28907 


lit 


M 


51474 


- 


2-h 


M 


51559 


- 


*4 





50129 


32271 


lit 


M 


51707 


- 


2^ 


M 


49422 


- 


ilif 


K 


56577 


- 


lit 


M 


51242 


-' 


2-g" 


M 


50481 


- 


*H 


B 


53655 


- 


2 


K 


60213 


3I44I 


2i 


M 


51225 


- 


i*i 


C 


50969 


30814 


2 


K 


57567 


30839 


ft 


A 


48382 


30459 


ill 


G 


50310 


33565 


2 


Px 


529H 


31 198 


*h 


M 


51666 


- 


Hi 


E 


50307 


29767 


2 


M 


52820 


- 


tk 


M 


51530 


- 


1 1 d 


J 


48953 




2 


M 


49164 


- 


*i 


M 


51296 


- 


I* 


K 


55S03 


31031 


2 


E 


51818 


27318 


A 


F 


48812 


- 


l 8 


C 


54447 


32334 


2 


P 


51684 


33104 


*\ 


F 


48894 


- 


T 7 

A 8 


D 


53100 


3 2 °74 


2 


P 


50834 


31878 


2i 


F 


49164 


31966 


I* 


N 


54004 


33610 


2 


N 


52127 


32461 


2{ 


F 


49290 


32163 


if 


Fxl 


52875 


35641 


2 


N 


51370 


32460 


2i 


P 


46866 


2824I 


if 


Fx3 


5336i 


35032 


2 


Fxl 


5201 1 


34702 


2i 


F 


47344 


29758 


I's 


P 


52505 


32312 


2 


C 


5"53 


29335 


2\ 


F 


48475 


28932 


I* 


E 


50880 


27100 


2 


D 


51 146 


28567 


2k 


F 


47428 


29941 


1 » 


D 


5 r 459 


27816 


2 


P 


49872 


29953 


2f 


F 


46446 


26333 


t% 


Px 


51762 


32261 


2 


Fx2 


50000 


36184 


3 


F 


47761 


264OO 


If 


M 


50363 




2 


Fx3 


50763 


33172 


3i 


F 


47014 


24591 


H 


A 


50584 


28713 


2 


A 


5017 1 


28983 


» 


F 


47000 


24961 


If 


F 


51039 


33067 


2 


F 


48596 


27634 


3f 


F 


46667 


23636 


if 


Fx2 


5II59 


3397o 


2 


F 


47812 


35864 


4 


F 


46322 


2343° 


If 


F 


49744 


356i5 


2 


F 


47569 


28792 











374 



APPLIED MECHANICS. 



TENSILE TESTS MADE SUBSEQUENTLY AT THE WATERTOWN 

ARSENAL. 

Here will next be given, in tabulated form, the results of a 
number of tensile tests made on the government machine at the 
Watertown Arsenal. 

The following tables of results on rolled bars, from the Elmira 
Rolling-Mill Company {mark L) and from the Passaic Rolling- 
Mills (mark S), are given in Executive Document 12, 47th Con- 
gress, 1st session, and in Executive Document I, /ffth Congress, 
2d session. 

SINGLE REFINED BARS. 



pq 

e 


as 

s 


c 

of <8 

aj ,c 
< .5 
"« £> 

C a 
.9 3 

u °" 

4) W 

CO 


Elastic Limit, in 
lbs., per Square 
Inch. 


Ultimate Strength, 
in lbs., per 
Square Inch. 




00 

- 

S.s 

3 


•1 ^ 
2 i 

I* 


Appearance of 
Fracture. 


Modulus of Elas- 
ticity at Load of 
20000 Lbs. per 
Square Inch. 




3 
. 

fa 


O us 


L 1 


3.06 


2850O 


52710 


18.4 


33-3 


95 


5 


26981450 




L 2 


3.06 


2950O 


53 6 30 


16.4 


36.0 


92 


8 


27826036 




L 3 


3.06 


290OO 


52090 


21.4 


34-6 


95 


5 


28419182 




L 4 


3.06 


290OO 


5 '440 


15.0 


20.3 


90 


10 


30888030 




L 5 


6.46 


2750O 


50500 


14.5 


27.6 


95 


5 


27826036 




L 6 


6.40 


2750O 


50530 


*7-3 


22.3 


70 


30 


271 18644 




L 7 


6.39 


270OO 


50200 


18.0 


22.5 


95 


5 


27444253 




L 8 


3-24 


- 


51667 


22.0 


36.0 


70 


30 


28318584 


Round. 


L 9 


3.20 


- 


50844 


16.3 


22.0 


15 


85 


27972027 


" 


L 10 


3.20 


- 


53062 


21.0 


40.0 


95 


5 


281 19507 


«« 


S 11 


3.08 


2850O 


48640 


*3-3 


24-3 


100 


Slightly 


27586206 




S 12 


3.08 


280OO 


50390 


16.9 


35-i 


100 





27586206 




s 13 


3-05 


2850O 


47050 


9.0 


22.0 


100 


• 


27874564 • 




S 15 


6.40 


260OO 


49700 


17.1 


19.2 


85 


15 


29906542 




S 16 


6.40 


240OO 


49280 


15-7 


17.7 


85 


J 5 


26490066 




S 17 


6.41 


2450O 


48740 


14-3 


17-3 


80 


20 


281 19507 




S 18 


3-*7 


2460O 


49680 


19-5 


32.0 


100 


Slightly 


27972027 


Round. 


S 19 


3-i7 


25870 


49338 


18.3 


38.0 


100 





29357798 


" 


S 20 


3-i7 


24920 


48864 


18.4 


37-o 


100 


Cinder 
at centre 


27729636 


«< 



DOUBLE REFINED BARS. 



375 



DOUBLE REFINED BARS. 





8 J 
< J 

§ 2 

u </> 

CO 


Elastic Limit, in 
lbs., per Square 
Inch. 


Ultimate Strength, 
in lbs., per 
Square Inch. 



00 

rt <2 

8J 

w 


•1 ^ 


Appearance of 
Fracture. 


Modulus of Elas- 
ticity at Load of 
20000 lbs. per 
Square Inch. 






"3 ^ 
u - 


L 201 


3.06 


290OO 


53560 


15-3 


37-9 


100 





27633 8 5 I 




L 202 


3-03 


30OOO 


5 2 650 


16.2 


20.6 


85 


15 


34042553 




L 203 


3.06 


32500 


53500 


16.5 


27.5 


100 





28 I 690I 4 




L 204 


3.06 


325OO 


54480 


154 


24.8 


100 





29090909 




L 205 


6-33 


27OOO 


5 I2 3° 


17.8 


24.2 


80 


20 


281 I9507 




L 206 


6-34 


27500 


50500 


17.6 


21. 1 


100 


Slightly 


29629629 




L 207 


6-34 


270OO 


51030 


21.4 


3i-9 


100 





27826086 




L 208 


3.20 


- 


50156 


22.7 


43-o 


100 


Cup- 
shaped 


2802IOI5 


Round. 


L 209 


3.20 


- 


49937 


22.6 


45-o 


100 


" 


2862254O 


" 


L 210 


3.20 


- 


50188 


19.9 


43-o 


100 


a 


28985507 


« 


S 211 


3-°5 


2950O 


5" 5o 


22.0 


3i-5 


100 





32989690 




S 212 


3-°5 


28500 


51110 


22.0 


36.1 


100 





25559105 




S .213 


3." 


29500 


51860 


22.5 


39-2 


100 





26446280 




S 215 


6.31 


27500 


50980 


19.-1 


23.6 


95 


5 


29357798 




S 216 


6.38 


27000 


50770 


20.7 


29.6 


100 





2826855I 




S 217 


6-33 


27000 


5 T 340 


19-3 


35-2 


100 





28070175 




S 218 


3-*7 


24610 


50631 


20.4 


41.0 


100 





2862254O 


Round. 


S 219 


3-17 


- 


50915 


25-5 


44.0 


100 


Cup- 
shaped 


2826855I 


" 


S 220 


3-17 


- 


50205 


237 


44.0 


100 


" 


28070175 


" 



The moduli of elasticity had not been computed in the 
report, but have been computed in these tables from the elon- 
gations under a load of 20000 lbs. per square inch in each case, 
as recorded in the details of the tests. 

In these reports are also to be found tensile tests of iron 
from other companies, as the Detroit Bridge Company, the 
Phoenix Company, the Pencoyd Company, etc. Some of these 



376 



APPLIED MECHANICS. 



tests were made to determine the effect of rest upon the bar 
after it had been strained to its ultimate strength. Some were 
made to determine the values of the modulus of elasticity 
of the same iron for tension and for compression ; and these 
were found experimentally to be almost identical, as was to 
be expected. For these tests the student is referred to the 
reports themselves ; and only certain tests on eye-bars of the 
Phcenix Company will be appended here. Some of them were 
tested only for modulus of elasticity, which are not calculated 
in the report, but are given here. 



Arsenal 
Num- 
ber. 


Outside 
Length, 
inches. 


Gauged 
Length, 
inches. 


Sectional 
Area, 
square 
inches. 


Modulus 

of 
Elasticity. 


Ultimate 

Strength, 

lbs., per 

Sq. In. 


Con- 
traction 
of Area 
at Frac- 
ture, %. 


5°9 


95-90 


70 


O.765 


25309000 






5io 


95-90 


70 


O.765 


27035000 






5 11 


67-75 


5° 


I.478 


22391000 


40600 


16.8 


5 12 


67-75 


5° 


15.090 


25075000 






5 J 3 


67.80 


5° 


I.940 


24727000 


39480 


13-9 


514 


118.75 


90 


J-959. 


2459OOOO 






5*5 


118.75 


90 


i-959 


25993000 






5i6 


96.05 


75 


2.954 


25388000 






5i7 


IO9.72 


75 


3.800 


25969000 






5i8 


96.05 


75 


* 5- io 3 


2333 8 000 


46720 


8.1 






Quite a number of tests of the iron of different American 
companies are to be found in the "Report on the Progress of 
Work on the Cincinnati Southern Railway," by Thomas D. 
Lovett, Nov. 1, 1875. 

For these the student is referred to the report named. 



SPECIAL MODULUS OF ELASTIC LTY. 2>77 



SPECIAL MODULUS OF ELASTICITY. 

In connection with Hodgkinson's experiment on a bar 50 
feet long, attention was called to the fact that the modulus 
of elasticity remained more nearly constant in wrought than in 
cast iron for loads below the limit of elasticity, and that, after 
passing this limit, the modulus of elasticity decreases rapidly, 
inasmuch as the strain increases very much more rapidly than 
the stress. 

This could be represented graphically by drawing a curve, 
having for abscissae of its several points the loads per square 
inch applied to the piece, and for ordinates the elongations or 
shortenings. Such a curve would, in the case of wrought-iron, 
be nearly a straight line up to the limit of elasticity, and then 
would rise very rapidly. 

If, now, we should plot another curve with the same abscissae 
as before, but having for ordinates the permanent sets under 
the respective loads, this curve would start from the axis of 
abscissae at the limit of elasticity, and rise rapidly from that 
point on. 

Again : if another curve be drawn, having the same abscissae 
as before, but having for ordinates in every case the differences 
of the ordinates of the other two curves, such a curve would 
represent, by its ordinate at any point, the recoil or the stretch 
of the piece over and above the permanent set ; and this 
ordinate would, in the absence of further experiment, appear 
to represent the amount it would stretch if the load were 
removed and immediately applied again, over and above that 
part of the stretch which did not disappear on the removal 
of the load. The modulus of elasticity computed by means of 
the elongation above described, considered as the elongation 
of the piece, has been given the name " special modulus of 
elasticity " by Col. Rosset of the arsenal at Turin ; and he has 
called attention to the fact, that in the case of wrought-iron, 



378 APPLIED MECHANICS. 

and also of steel, this special modulus of elasticity is nearly 
constant, even though the load applied be far above the limit 
of elasticity. 

This can be graphically shown from the fact that the third 
of the above-mentioned curves, if plotted, will be nearly a 
straight line almost up to the point of fracture in the case of 
wrought-iron and steel. 



EXAMPLES. 

i. Plot the curves referred to above, for Hodgkinson's test of a 
wrought-iron rod 50 feet long, recorded on p. 353. 

2. Do the same for the table of experiments on cast-iron bars 10 
feet long, recorded on p. 339. 



WROUGHT-IRON BOILER-PLATE. 

Some tests of iron boiler-plate have already been quoted ; 
viz., Kirkaldy's. Some tables of tests made by Mr. C. B. Rich- 
ards, and recorded in the "Transactions of the American So- 
ciety of Civil Engineers," vol. ii., will next be given. 

In these tables, L denotes that they were pulled lengthwise 
of the fibre, and C crosswise. 

Long specimens are those whose smallest area of cross- 
section had a length of from 3 to 5 inches. 

Short specimens are those whose smallest area of cross- 
section had no length. 



WR O UGHT-IR ON B OILER-PL A TE. 



379 





tO tO tO tO 


4» 4* 


to >-> 


to ii 


oj oj 


to to 


4u. 4i. 4^. to 4>- OJ OvOo On 


No. of 




















Specimen. 


m 


m a 


a 


a 


a 


a 


H 


a a a 






.. cr. £ p 


£ p 


n> 


o 


n> 


3* 


P £P £££££P 




n 


~CTO. ^ 


^ 


3 


3 


3 


o 


"-< ^ •-< 




o 

3' 


° in 


m 


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(A 


en 


3 
oo 


3 


in en in 




3 s p 


c-t- 

£ P 


£M. 


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" ^ 




ti £p£££££P 








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►n 


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papapapa 


Direction of 
Lamination. 


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w O « O « 




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00 


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o'o ff 
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vj 


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ft 




on to 


VO 


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00 


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38o 



APPLIED MECHANICS. 



Also the following results on rivet-iron manufactured by Burden 
& Sons : — 



Test 
No. 


u c 

a v 

— 


11 

6 


■£ J* 

to u 
c s 

•3~ 


Tensile Strength per 
Square Inch. 


Limit of 

Elasticity, 
lbs., per 
Sq. Inch. 


Con- 
traction 
of Area, 
pr. cent. 


Ulti- 
mate 
Elonga- 
tion, pei 
cent. 


Original 
Area. 


Fractured 
Area. 


3 2 4 


Long, 


0.620 


5-° 


49206 


93459 


3M5° 


47 


33 


325 


" 


0.619 


5-o 


49734 


94218 


26580 


47 


32 


326 


<( 


0.619 


5-° 


49535 


93774 


28240 


47 


29 


327 


«« 


0.619 


5-° 


49202 


97434 


26570 


49 


32 


332 


<« 


0.7 1 1 


S-o 


49345 


1 00000 


26460 


50 


34 


333 


« 


1. 000 


5-o 


49974 


86454 


- 


42 


28 


334 


" 


1. 000 


5-° 


49924 


91 186 


- 


45 


29 


335 


(I 


1. 000 


5-o 


49784 


93430 


- 


47 


26 


328 


Short, 


0.614 


- 


62230 


90620 


- 


30 


- 


3 2 9 


" 


0.616 


- 


61670 


88040 


- 


3i 


- 


33o 


« 


0.616 


- 


60840 


85358 


- 


29 


- 


33* 


« 


0.615 


- 


60407 


87941 


- 


3 1 


- 


336 


« 


1.002 


- 


63840 


88783 


- 


28 


- 


337 




1. 000 


- 


63547 


88025 


- 


28 


— 



Perhaps it ought to be stated that the boiler-plate tested 
and recorded in the first of these tables is boiler-plate of good 
quality in every case. 



BRANDS OF PLATES. 

The only safe way, in ordering plates for a boiler, as in or- 
dering iron for any construction, is to prescribe the tests which 
it shall stand, as the same brands are used by different makers 
to denote very different qualities of iron. 

As a rule, better iron will be obtained by purchasing at a 
mill where only superior qualities are manufactured, rather than 
at mills where all qualities are made. 



BRANDS OF PLATES. 38 1 

In England the brands are very various, but are often graded 
"Best," "Best Best," and "Treble Best," the "Best" being the 
poorest of the three. 

In the United States these brands are not used, and the 
usage varies with different mills. Some years ago the brands 
manufactured were as follows : — 

i°. Tank-Iron, the lowest grade, not suitable for use in the 
shell of a boiler, and including all iron too poor for this purpose. 

2°. C. No. I. — This used to denote iron made by using only 
charcoal in the blast-furnace ; and it furnished a good grade of 
iron, suitable to use in those parts of the shell of a steam-boiler 
which do not come in contact with the fire, and which do not 
have to be flanged. 

3 . C. H. No. I, or charcoal-hammered No. 1, which was 
also a charcoal iron, but was obtained by more working than 
C. No. 1, and was hence a better grade of iron. 

4 . Fire-box Iron, suitable to use for sheets exposed to the 
fire. 

5°. Flange Iron, or such as will flange without cracking. 

These brands partly remain to-day, but they no longer bear 
the same meanings. Indeed, the name " C. H. No. 1 " is, in 
many mills, given to iron in the manufacture of which no char- 
coal has been used. 

There are a number of mills that use the following brands 
for boiler-plates ; viz., — 

Tank- Iron. Extra Flange. 

Refined Iron. Shell Fire-Box. 

Shell Iron. ' C. H. No. 1 Fire-Box. 
C. H. No. 1. Extra Flange Fire-Box. 

Nevertheless, if we buy any one of these grades at different 
mills, we have no certainty of obtaining the same quality of 
iron ; and, moreover, there are other mills that use different 
brands. " Refined iron," for instance, is a term that includes a 



382 APPLIED MECHANICS. 

large number of qualities, beginning at much too low a grade to 
put in a boiler-shell. 

The only sure way to secure good iron is to prescribe the 
tests it shall stand : i.e., the tensile strength, which should be 
over 45000 or 46000 lbs. ; the limit of elasticity, which should 
be as much as 27000 or 28000 lbs. ; the contraction of area at 
fracture, which should be at least 30 per cent. It should also 
stand bending double cold, red-hot, and at a flanging heat. It 
would be well to test also the soundness of the plates by 
punching them. 

BRANDS OF BARS AND SHAPES. 

For these, each mill has its own peculiar brands, to which it 
attaches its own signification. The only way to secure iron of 
whose quality we are sure, is, therefore, to require that it shall 
stand the suitable tests before being accepted. 

COMPRESSIVE STRENGTH OF WROUGHT-IRON. 

As far as the compressive strength of short cylinders of 
wrought-iron is concerned, we have hardly any direct experi- 
mental evidence ; because the iron, especially if it is soft, will 
gradually flatten out, and not give way by suddenly breaking. 
The evidence furnished by Fairbairn's experiments on trans- 
verse strength of wrought-iron goes to show that the tensile 
and the crushing strength are nearly equal, though the earlier 
experiments of Fairbairn had given a less value for the crush- 
ing than for the tensile strength ; and for a long time the 
crushing-strength of wrought-iron has been given as 36000 lbs. 
per square inch, which value is still to be found in Gordon's 
formula for the strength of wrought-iron columns. 

Leaving this matter to be explained under the head of 
"Transverse Strength of Wrought-iron," we will next proceed 
to consider the strength of wrought-iron columns. 



WROUGHT-IRON COLUMNS. 383 

§ 223. Wrought-Iron Columns. — Until a very recent 
date, we have had no experimental knowledge on this subject 
beyond the experiments of Hodgkinson, which have furnished 
the constants for Hodgkinson's, and also for Gordon's, formula, 
as already given in § 209 and § 211. 

These formulae have been in very general use, and it is only 
very recently that we have been able to test their accuracy by 
tests on full-size wrought-iron columns. The disagreement of 
the formulae already referred to, with the results of the tests, has 
led to the proposal of a large number of similar formulae, each 
having its constants determined to suit a certain definite set of 
tests, and hence all these formulae thus proposed must be classed 
as empirical formulae, and can only be applied with safety within 
the range of the cases experimented upon. 

A number of these will now be enumerated: and then will 
follow tables of the actual tests, which furnish the best means 
of determining the strength of these columns ; and it would 
appear that it is these tables themselves which the engineer 
would wish to use in designing any structure. 

On the 15th of June, 1881, Mr. Clark, of the firm of Clark, 
Reeves, & Co., presented to the American Society of Civil 
Engineers a report of a number of tests on full-size Phoenix 
columns, made for them at the Watertown Arsenal, together 
with a comparison of the actual breaking-weights with those 
which would have been obtained by using the common form of 
Gordon's formula for wrought-iron, 

P _ 36000 



36ooop 2 



where P = breaking-weight in lbs., A = area of section in 
square inches, / = length in inches, p = least radius of gyra- 
tion in inches. The table is as follows : — 



3§4 



APPLIED MECHANICS. 



Total 

Ultimate 

Strength, in 

lbs., by 

Gordon's 

Formula. 


VO 0\nO\NfOlONi- ►* OVOVO t^w -rf-N 1-^ 

Tfm« w ro rf u-> On O 0*0*000 t)- N M rOO 
t-i^Oi-iooOOM-it^ r^O oooo O OfOM in 

OfONOONHKOMMOvO^NrOONM 

rOfO^i'+NN r-»00 0\0»h OWN roo ro ro 
rorororororororororo^-TfrfTt-Ti-Tf^-Tj- 


1 1 1 1 


to 

G 
<U 

CO 
V 

a 
B 

5 


«5 JJ . 

P<rt-fS 

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WROUGHT-IRON COLUMNS. 385 

On the other hand, if the breaking-weights of these columns 
were calculated from the formula 

P 50000 
~a T 2 — ' 

i+— - 

36000^ 

the results would be much nearer the actual results, though not 
very near. The Phcenix Company, in their handbook, give, 
for the strength of their columns, 

P 50000 



3Qoo^ 2 



where h = diameter in inches. 

The set of tests above referred to, gave rise to considerable 
discussion, and the proposal of several modified forms of Gor- 
don's formula, as well as other formulae, which should agree 
more nearly with the results of the tests. 

Mr. Bouscaren proposes the formula 

P _ 38000 

A = 



1 + 



IOOOOOp 2 



and claims that this furnishes a very good degree of coincidence 
with the facts. 

Mr. Theodore Cooper proposes, instead, a modified form of 
Gordon's formula, as follows : — 

P f 

For square-ended columns . . . — = i. . 

A •+(--)• 



i P 



18000 



386 APPLIED MECHANICS. 






For pin-ended columns . 



For round-ended columns 



And he gives, for the values of f, 



p _ / 






■+(-«)■ 



18000 



A 



\p/ 18000 



For Phoenix columns f = 36000. 

For American Company's columns. . . f = 30000. 
For box and open columns f = 31000. 

He deduces these values of /from some tests made in 1875 
by Mr. Bouscaren, combined with those, already referred to, 
made at the Watertown Arsenal. The box and open columns 
were made of channel-bars and latticing. The tables or dia- 
grams presented to justify all these formulae will not be given 
here, but any one can find them in the " Transactions of the 
American Society of Civil Engineers" for 1882. Mr. C. E. 
Emery proposed, at the same meeting, the formula 

P_ _ 355 o6 3 + 3095°* 
A ~ x + 6.175 

for the Phoenix columns tested, where x = number of diameters 
in the length. 

Professor Burr proposes 



P 

A 



42000 1 -h log^i -h jj\ 



50000 p 2 



WROUGHT-IRON COLUMNS. 387 



Professor Merriman proposes 



P _ 37200 

a r 

1 4- 



158500^ 



The above are the formulae evoked from the discussion on 
the set of tests of Phoenix columns above referred to. 

In the " Report of Progress of Work of the Cincinnati 
Southern Railway," by Thomas D. Lovett (Nov. 1, 1875), is 
given an account of some tests made on full-size wrought-iron 
columns. He concludes that the formula 

P _ f 

A r J J 2 

E -f- a— 

P 2 

will give results agreeing fairly well with the facts, provided we 
use 

For flat ends a' = , 

36000 

For flat at one end and round at the other, a r = 



36000' 

and for f we put the crushing-strength of the iron per square 
inch. 

On the other hand, Professor Burr deduces from these same 
tests the following formulae (see Burr's " Elasticity and Resist- 
ance of Materials," p. 441) : — 

KEYSTONE COLUMNS. 

Flat ends (swelled columns) . . . — = 

A 1 -I 

' 1830OP 2 



388 ' APPLIED MECHANICS. 



Flat ends straight j ^ [ . . \ ^ = ^ * 9S ™ t 

1 18300/) 2 



Pin ends swelled .... . . . -j = j- 2 

1 +— 

15000P 2 



SQUARE COLUMNS. 
171 . A P 3900O 

Flat ends -7 = -^ — 

1 + 35000P 2 

P 39000 
Pin ends -% = 75 — 

i+— 

J7000/0 2 

PHGENIX COLUMNS. 
171 * A P 42000 

Flat ends -7 = 7^ — ■ 

^ 1 +— i 

50000P 2 



r, j j ^ 4200O 

Round ends -; = jz — * 

i+— 

12500/0 2 



P 42000 

Pin ends . A = 7^ 

1 H 

227OO/0 2 



WROUGHT-IRON COLUMNS. 389 

AMERICAN BRIDGE COMPANY'S COLUMNS. 

P 360OO 



Flat ends * 



1 + 



Round ends * 



46000P 2 
P 36000 



1 + 



Pin ends 



1150OP 2 
P _ 36000 

3 "i+ /2 



21500P 2 



Professor Burr has also determined the values of f and a in 
the formula 

P f 

a = "F 

. 1 + of- 

P 2 

which should agree most nearly with the different sets of 
experimental results. He has also done the same for the val- 
ues of a and n in the formula 

P (p\ n 



al j r 



this latter being a modification of Euler's formula; and he 
claims that the latter gives results agreeing better with experi- 
ment than the former. 

The tests made at the, Watertown Arsenal will next be 
given, together with cuts showing the form of the columns ; 
these being taken from the government report, Executive 
Document 12, 47th Congress, first session. 



39° 



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TRANSVERSE STRENGTH OF WROUGHT-IRON. 397 

conclusions ; and the formulae proposed by the different persons 
who have deduced formulae to cover special sets of values may- 
be advantageously used for cases intermediate between those 
that have been experimented upon. 

§ 224. Transverse Strength of Wrought-Iron. — In all 
probability wrought-iron owes its extensive introduction into 
construction as much or more to the efforts of Sir William 
Fairbairn than to any one else ; and while he was furnishing 
the means to Eaton Hodgkinson to make extensive experiments 
on cast-iron columns, and while he made experiments himself 
on cast-iron beams, which were in use at that time, he also 
carried on a large number of tests on beams built of wrought- 
iron, more especially those of tubular form, and those having 
an I or a T section, and made of pieces riveted together. In 
his book on the "Application of Cast and Wrought Iron to 
Building Purposes " he gives an account of a large number 
of these experiments, including those made for the purpose of 
designing the Britannia and Conway tubular bridges, a fuller 
account of which will be found in his book entitled "An Ac- 
count of the Construction of the Britannia and Conway Tubular 
Bridges." In the first-named treatise. he urges very strongly 
the use of wrought-iron, instead of cast-iron, to bear a trans- 
verse load. 

Fairbairn tested a number of wrought-iron built-up beams 
with different thicknesses of upper and lower flange. The first 
results he obtained showed, that, unless the upper plates were 
made very much thicker than the lower, the beam would invari- 
ably give way by crippling at the top. 

At first he concluded that the tensile strength of wrought- 
iron is greater than its compressive strength, and it has often 
been so regarded. 

Subsequent experiments made by him, however, showed, 
that, if the iron of the upper flange were distributed in the form 
of cells, the areas of the cross-sections required in order to 



;g8 



APPLIED MECHANICS. 



render the beam equally liable to give way by tearing or by 
crushing, were as 12 to n, or nearly equal; thus tending to 
show that the tensile and compressive strengths of wrought- 
iron are nearly equal, and that the reason of the crippling in 
the first experiments was, that the iron between the rivets 
acted like a column, and bent, instead of bringing into play the 
entire compressive strength of the iron. 

A summary of some of these first experiments, the details 
of which are in his " Application of Cast and Wrought Iron to 
Building Purposes," is given in the following table, taken from 
p. 116 of his book : — 















Ulti- 






No. of 
Exper- 
iment. 


Distance 
between 
Supports. 


Depth, 

in 
inches. 


Width, 

in 
inches. 


Thick- 
ness of 
Top 
Plate. 


Thick- 
ness of 
Bottom 
Plate. 


mate 
Deflec- 
tion, 
in 
inches. 


Breaking- 
weight, 
in lbs. 


Manner of Failure.. 




ft. in. 
















14 


17 6 


9.60 


9.60 


O.075 


O.074 


1. 12 


3733 


Compression. 


14 a 


17 6 


9.60 


9.60 


O.252 


O.075 


I. IO 


8273 


Tension. 


15 


17 6 


9.60 


9.60 


O.076 


O.143 


O.94 


3788 


Compression. 


15 a 


17 6 


9.60 


9.60 


O.143 


O.076 


I.76 


7148 


Compression. 


16 


17 6 


18.25 


9-25 


O.I49 


O.269 


I.03 


68l2 


Compression. 


16 a 


17 6 


18.25 


9-25 


O.269 


O.149 


i-73 


I2I88 


Compression. 


17 


24 


I5.OO 


2.25 


O.260 


O.260 


1.61 


I3I20 


Compression. 


18 


18 


l 3- 2 S 


7-5° 


O.143 


O.143 


I-3 1 


10880 


Compression. 


25 


11 


8.00 


1. 00 


O.282 


O.Il6 


075 


7146 


Compression. 


29 


19 


15.40 


7-75 


O.II5 


O.180 


i-S9 


22469 


Side plate tore. 



Very few tests have been made which claim to give the 
modulus of rupture of wrought-iron, for the reason that wrought- 
iron will bend very much before breaking when it is subjected 
to a transverse load. 

Some experiments made on small pieces claim to show a 
modulus of rupture considerably in excess of the tensile or 



TRANSVERSE STRENGTH OF WROUGHT-IRON. 399 

compressive strength, while the few experiments that have 
been made for this purpose on full-size beams have given a 
modulus of rupture about equal to the tensile or compressive 
strength of the iron per square inch. 

Of the experiments on deflection, however, we generally 
have a very uniform result shown for the modulus of elasticity, 
both from small and from large beams. As to experiments on 
large beams, besides those of Fairbairn, already referred to, we 
have : — 

i°. Those made at the Watertown Arsenal by Mr. William 
Sooy Smith and by Col. Laidley. 

2°. Some tests made for the Phcenix Iron Company, and 
recorded in their published handbook. 

3 . Some tests made for the New-Jersey Iron and Steel 
Company, and recorded in their handbook. 

Those made at the Watertown Arsenal are recorded in 
Executive Document 23, 46th Congress, second session, and 
were made merely on the deflections under moderate loads, and 
on the elastic limit. From those made on the deflections under 
moderate loads, the moduli of elasticity are deduced ; and they 
agree very closely with the moduli of elasticity that are deduced 
from experiments on the tension of wrought-iron. Because 
those on the elastic limit do not agree with the tensile elastic 
limit of the outside fibre, does not seem to the writer to justify 
Mr. Smith in looking for some other element in the resistance 
besides the direct tensions and compressions recognized in the 
common theory. 

The table of these results will now be given, and then the 
results obtained from some of the Iron Company's beams. 

The experiments were made on T-beams with the flange 
upwards. 

These tests give a higher modulus of elasticity than would 
seem advisable to use in the present state of our knowledge of 
the subject. 



400 



APPLIED MECHANICS. 



g 

a 

pq 
in 

o 

6 


Depth, 

in 
inches. 


Moment 

of 
Inertia. 


No. 

of 

Trials. 


Span, 

in 
feet. 


Elastic Limitt 


/- 


E. 


No. 

of 

Trials. 


Weight 
applied. 


One- 
half 
Weight 
of 
Span. 


Ultimate 

Strain in 

Outer 

Fibre. 


Co-efficient 

of 
Elasticitv. 


x 


15 


536-56 


7 


20 


6 


21225 


486 


I82IO 


3432850O 


2 


io£ 


221.86 


24 


22 


19 


18647 


479 


29866 


260994OO 


3 


EOJt 


174-75 


14 


22 


14 


12741 


383 


26097 


289030OO 


4 


10^ 


174-75 


17 


22 


15 


13798 


387 


28721 


29915000 


5 


ioi 


154.90 


9 


21 


5 


1 1476 


322 


2519I 


31 16700O 


6 


9 


106.53 


14 


22 


11 


9973 


3 2 7 


28713 


3262840O 


7 


9 


106.53 


12 


13 


11 


17701 


192 


29478 


354090OO 


8 


9 


106.53 


5 


II 


4-5 


21000 


163 


29500 


30720OOO 


9 

10 


8 
8 


62.34 
62.34 


8 
6 


22 
22 


» 


6962 


241 


30508 


( 31859000 
( 29270200 


ii 


8 


62.34 


9 


14 


8 


12237 


165 


33957 


26863750 


12 


8 


62.34 


9 


14 


7 ■ 


1 096 1 


165 


30000 


288167 IO 


13 


7 


44.12 


6 


22 


*1 








f 32834200 
•| 308360CO 
[ 33333800 


14 


7 


44.12 


5 


22 


4 


5346 


236 


27413 


*5 


7 


44.12 


5 


22 


4 J 








16 
17 


7 
7 


44.12 
44.12 


7 
7 


14 
14 


a 


9704 


173 


32575 


( 3OI2870O 
I 30099300 


18 


6 


24.58 


4 


22 


4 


4189 


175 


34393 


33799500 


19 


6 


24.58 


6 


14 


5 


6364 


114 


33207 


34057600 


20 


6 


24.58 


6 


14 


5 


7631 


114 


365 6 4 


32453400 


21 


5 


12.85 


6 


II 


5 


535i 


60 


3474o 


340648CO 


22 


5 


12.85 


6 


II 


5 


5877 


60 


38116 


3163180O 


2 3 


5 


12.85 


5 


II 


5 


5877 


60 


38116 


3097600O 


24 
25 


4 
4 


7-42 
7.42 


6 

5 


» 


if 


4189 


54 


3775° 

*30II7 


( 36064400 
( 32687000 

t3II2S26o 




Extr 


ernes . 










• 1 


25191 
38116 


2609940O 
36664400 



* Average of 24 tests. 



f Average of 25 tests. 



TRANSVERSE STRENGTH OF WR OUGHT-IRON. 



4OI 



The New-Jersey Iron and Steel Company give 12000 lbs. 
per square inch as the safe extreme fibre stress to be used in 
computing the safe load on any of their beams ; and they claim, 
that, in so doing, they are using a factor of safety of four. 

The rule which they give for computing deflections, on the 
other hand, corresponds to a modulus of elasticity of about 
21000000 lbs. per square inch. 

The Phoenix Company also give 12000 lbs. per square inch 
for extreme fibre stress, and claim, that, in so doing, they are 
using a factor of safety of about four. They give, for modulus 
of elasticity to be used in computing deflections, 24000000 lbs. 
per square inch. 

Both companies claim to have had some of their beams 
tested, and that such tests support them in recommending the 
values s;iven above. 

Some of the tables of tests given in these handbooks will 
be placed here. The following table of tests, made for the 
New-Jersey Iron and Steel Company by a United-States gov- 
ernment engineer, is given in their handbook ; and it is stated 
that they are a few of the tests selected at random. 

TESTS OP BEAMS OF NEW-JERSEY IRON AND STEEL COMPANY. 



Size of Beam. 


Clear 
Span, 


How Loaded. 


Safe 
Load, in 
lbs., as 
given by 


Load 
actually 


Ratio 

of 
Actual 


Effect on 
Beam. 


c 
J -a 

SI § 


c 

3 

.2 1? 

u O 
U iJ 

"S.o 


"o '-3 

\3 C 

t/i u 

j2 pa 




in feet. 




the Com- 
pany's 
Tables. 


applied. 


Safe 
Load. 


"3 "2 

5 = 


< 


Limit 
Permai 

begins. 


6-in. light . 


12.00 


At centre, 


2608 


1 1000 


4-3 


Failed . 


O.27 


O.30 


7000 


6-in. light . 


n-93 


<< << 


2624 


I70OO 


6.5 


" 


O.27 


0.15 


I IOOO 


9-in. heavy, 


14-93 


«« << 


6330 


320OO 


5-i 


«« 


O.28 


O.16 


22000 


15-in. light . 


21.00] 


Uniformly 
distributed, 


j 25188 


90OOO 


3-6 [ 


Deflected 
2.7 in. 


J O.42 


O.36 


• " 



402 



APPLIED MECHANICS. 



The following tables are given by the Phoenix Company as 
the results of tests made at different times of their rolled 
beams : — 



7-inch Beam. 60 lbs. per Yard. 


Clear Span, 21 Feet. 


Centre Load, 


Deflection, 


Increase, in 


in lbs. 


in inches. 


inches. 


2000 


O.468 


_ 


3000 


0-743 


O.275 


40OO 


I.020 


O.277 


5000 


I.298 


O.278 


O 


O.O29 


- 


6000 


I.578 


O.280 


O 


O.030 


- 


7000 


I.887 


O.309 


O 


O.060 


- 


SOOO 


2.3OO 


O.413 


O 


O.183 


- 


9000 


3- 54o 


I.24O 


95OO 


5.298 


I.758 


1 0000 


( Broke 
I yielding 


slowly, 
at top. 



9-inch Beam. 187 


lbs. per Yard. Clear 




Span, 


21 Feet. 




Centre 


Deflection, 


Increase, 




Load, in 


in 


in 


Remarks. 


lbs. 


inches. 


inches. 




2000 


0.228 


_ 




4000 


O.474 


O.246 




6000 


O.72O 


O.246 




8000 


0962 


O.242 




1 0000 


1. 201 


O.239 




O 


O.O48 


- 




12000 


1.432 


O.23I 




O 


O.05O 


- 




I3OOO 


T.580 


O.I48 




O 


0.1 17 


- 




I4OOO 


I.863 


O.283 







O.269 


- 




160OO 


3-256 


1-393 


f Side de- 


17000 


5-233 


1.997 


-j flection 
[ begins, 
r Beam 


17500 


5.602 


0.369 


\ yields 
[ slowly. 



TRANSVERSE STRENGTH OF WROUGHT-IRON. 



403 



9-inch Beam. 150 lbs. per Yard. 
Clear Span, 14 Feet. 


Centre Load, 


Deflection, 


Increase, 


in lbs. 


in inches. 


in inches. 


5608 


O.I02 


_ 


6720 


O.I26 


O.O24 


7840 


O.I48 


0.022 


8960 


O.I70 


0.022 


IO080 


O.I92 


O.022 


1 1 200 


O.214 


0.022 


I2320 


O.239 


O.O25 


13440 


O.261 


0.022 


14560 


O.287 


O.O26 


15680 


O.3IO 


O.O23 


16800 


0.336 


O.O26 


17920 


o-359 


O.O23 


I904O 


0.382 


0.023 


20160 


0.409 


O.O27 


21280 


0-435 


O.O26 


224OO 


0.458 


O.O23 


23520 


0.487 


O.O29 


2464O 


0.516 


O.O29 


25760 


0-543 


O.O27 


26880 


0.572 


O.O26 


280OO 


0.600 


O.O38 


2912O 


0.633 


0.033 1 1 i 


29120 


0.682 


0.049 1«^ 
13 


O 


0.082 


( Weight 
( removed. 



15-inch Beam. 200 Lbs 


. per Yard. 


Clear Span, 14 Feet. 


Centre Load, 


Deflection, 


Increase, 


in lbs. 


in inches. 


in inches. 


6720 


O.048 


_ 


8960 


O.060 


O.OI2 


1 1 200 


O.073 


O.OI3 


13440 


O.090 


O.OI7 


I5680 


O.IO5 


O.OI5 


17920 


0.I20 


O.OI5 


20l6o 


O.134 


O.OI4 


224OO 


O.148 


O.OI4 


2464O 


0.l6l 


O.OI3 


26880 


O.178 


O.OI7 


29120 


O.191 


O.OI3 


31360 


O.206 


O.OI5 


33609 


0.222 


0.0l6 


35 8 40 


O.234 


O.OI2 


38080 


O.246 


O.OI2 


40329 


O.258 


O.OI2 


42660 


O.27I 


O.OI5 


44800 


O.287 


O.Ol6 


47040 


°-305 


O.OlS 



Weight removed; permanent set, 
0.016. After lapse of 1 hour, the 
load of 15 tons was replaced, and 
caused a total deflection of 0.222 
inches, as before. 



404 



APPLIED MECHANICS. 





12-inch Beam 


125 lbs. per 


Yard. Clear 


Span, 27 Feet. 


Centre Load, 


Deflection, 


Increase, 


Centre Load, 


Deflection, 


Increase, 


in lbs. 


in inches. 


in inches. 


in lbs. 


in inches. 


in inches. 


6720 


O.691 


_ 


15680 


1.800 


O.17O 


7840 


O.821 


O.I30 


16800 


I.976 


O.I76 


8960 


O.948 


O.I27 


17920 


2.228 


O.252 


IO080 


1. 06 1 


O.II3 


19040 


2-455 


O.227 


1 1 200 


1. 186 


O.125 


20l6o 


2.742 


O.287 


I2320 


I.328 


O.I42 


20720 


2.900 


O.158 


13340 


I.466 


O.I38 


20720 


2.965 


O.065 


I4560 


1-630 


O.164 









Last load left on 15 minutes, deflection increasing to 2.965. 

§225. Steel. — Steel is usually defined to be a compound 
of iron with a small percentage of carbon, this percentage 
varying from a very minute quantity up to one and a half, or at 
most two, per cent. Steel is made in one of the three follow- 
ing ways ; viz., — 

i°. By adding carbon to wrought-iron. 

2°. By removing carbon from cast-iron. 

3 . By melting together cast and wrought iron in suitable 
proportions. 

These three processes are, as a rule, represented respectively 
by the following three kinds of steel : — 

i°. Crucible steel, 

2°. Bessemer steel, 

3 . Open-hearth, or Siemens's Martin steel ; 
these being the three chief kinds that are made on the com- 
mercial scale. 

Crucible Steel. — This should always be, and is by good 
makers, made by re-melting blister steel in crucibles ; the lat- 
ter being made by the cementation process, in which bars of 
wrought-iron are heated in contact with charcoal until they 
have absorbed the necessary amount of carbon. 



STEEL. 405 



Crucible steel is used for the finest cutlery, tools, etc., and 
wherever a very pure and homogeneous quality of steel is 
required. 

Bessemer Steel is made by decarbonizing cast-iron by forcing 
a powerful blast through a mass of melted cast-iron, thus re- 
moving the greater part of its carbon, and then adding a small 
quantity of some very pure cast-iron which is rich in carbon, thus 
bringing up the percentage of carbon to the required amount. 

Open-hearth Steel is made by fusing a charge consisting of 
the suitable proportions of cast-iron with wrought-iron scrap, 
or with Bessemer steel scrap. 

Bessemer and open-hearth steel contain more impurities 
than crucible steel ; but they are very much cheaper, and are 
just as suitable for many purposes. It is only in consequence 
of their introduction that steel can be extensively used on the 
large scale, as crucible steel would be too expensive for many 
purposes. 

Steel is also made by puddling and by other processes. 

Steel, unlike wrought-iron, is fusible ; unlike cast-iron, it can 
be forged ; and, with the exception of the higher grades, it 
can be welded, the welding of high-grade steel in large masses 
being a very uncertain operation, though small masses can be 
welded by taking proper care. 

The special characteristic, however, is, that, with the ex- 
ception of the lowest grades, when raised to a red heat and 
suddenly cooled, it becomes hard and brittle, and that, by sub- 
sequent heating and slow cooling, the hardness may be reduced 
to any desired degree. The first process is called hardening, 
and the second tempering. 

Pure wrought-iron cannot be hardened by this means ; and 
cast-iron can be hardened, but cannot be tempered. 

Case Hardening is a process by which the outer coating of 
wrought-iron is turned into steel by heating it to a red heat in 
contact with bone-dust or some animal matter. This process 
gives the iron a hard surface combined with toughness. 



406 APPLIED MECHANICS. 

Whereas the hardening element of steel should be only- 
carbon, and whereas other substances should be absent as far 
as possible in the best steel, nevertheless phosphorus, silicon, 
and manganese, when present in small quantities, all have a 
hardening effect ; and all these ingredients, and often sulphur, 
are generally found in Bessemer and open-hearth steel, sul- 
phur, silicon, and phosphorus coming from the ore, the fuel, 
and the flux, and manganese being necessarily added, partly 
to counteract the effect of sulphur, partly, by its affinity for 
oxygen, to absorb any oxygen in the interior of the mass, and 
thus decrease the porosity, and partly to enable the steel to be 
welded by preventing the rapid oxidation of the surfaces at a 
high heat. 

When Bessemer steel" was first introduced, and for some 
time thereafter, it was chiefly used for rails. It is only since a 
more recent date that Bessemer and open-hearth steel have 
been made sufficiently homogeneous and reliable for use in 
construction generally ; but at the present day it is displacing 
wrought-iron in many instances, as being more reliable, and it 
is likely to displace it even more. 

In the construction of boilers, bridges, trusses, beams, etc., 
it will not do to use the higher grades of steel, but only the 
milder and more ductile kinds : thus steel with a tensile strength 
of more than 80000 or 90000 lbs. per square inch is generally too 
hard to use in construction ; and for steam-boilers, if its strength 
exceed about 65000 lbs., it is liable to be too hard. It should 
also show a large percentage of contraction of area, as 30 per 
cent or upwards. Such steel contains but little carbon, gen- 
erally not more than one-half per cent. 

While there are a few isolated cases where it is claimed by 
some that the structure of iron or steel may be changed from 
fibrous to crystalline without over-heating, the greater part of 
the evidence tends to show, that, whenever crystallization has 
taken place, it has occurred at a temperature above a welding- 
heat ; and in a great many instances where cold crystallization 



STEEL. 407 

has been claimed, it has been found on investigation that the 
piece has some time been over-heated. 

Welding is a much more difficult operation in steel than in 
iron, as (i°) there is always danger of over-heating, and (2 ) the 
metal does not unite as readily at a welding-heat ; hence, in 
high grades of steel, welding is almost an impossibility, espe- 
cially with large masses. 

The injury done to steel plates by punching is greater than 
that done to iron plates : this injury can, however, be removed 
by annealing. Steel requires greater care in working it than 
iron, whether in punching, flanging, riveting, or other methods 
of working ; otherwise it may, if over-heated, burn, or receive 
other injury from careless workmanship; 

In regard to the term "temper," it should be observed, that 
by the steel-maker it is used to denote the percentage of carbon 
in the steel, a higher temper corresponding to a higher per- 
centage of carbon. On the other hand, the term "temper" in 
common parlance refers to the degree of hardness as deter- 
mined by tempering the steel. 

The brands of steel are determined by each maker for him- 
self, there being no uniformity in this regard. 

The chemical composition of steel is one important element 
in its resisting properties ; but, on the other hand, the mode of 
working also has a great influence on the quality. 

The only means of securing good steel is, to prescribe the 
tests which it shall stand, and to reject all that does not fulfil 
the requirements. 

Thus, good boiler-plate should have an ultimate strength 
of 55000 to 65000 lbs. per square inch, a limit of elasticity of 
about 30000 pounds, a contraction of area at fracture of about 
30 per cent. It should not crack on (i°) being bent double cold, 
(2 ) at a red heat, (3 ) at a flanging-heat, and it should suffer 
but little injury by punching. 

For other purposes, as in trusses, etc., it should be able to 
stand, without injury, the trials to which it has to be subjected 



408 APPLIED MECHANICS. 

in construction, as bending, punching, riveting, etc. An ac- 
count of the manner of applying such tests to angle irons, 
I-beams, etc., will be found in " Use of Steel for Constructive 
Purposes," by J. Barba. 

Effect of Temperature upon the Resisting Properties of Iron 
and Steel. — The question of the effect of high and of low 
temperatures upon the resisting properties of iron and steel 
has been investigated by a number of different experimenters 
with seemingly discordant results. The following are the prin- 
cipal experimenters upon this subject : — 

Sir William Fairbairn : Useful Information for Engineers. 
Committee of Franklin Institute : Franklin Institute Journal. 
Knutt StyfTe and Christer P. Sandberg : Iron and Steel. 
Kollman : Engineering, July 30, 1880. 
Massachusetts Railroad Commissioners : Report of 1874. 

The existing evidence on the subject has been very care- 
fully collated, and compared by graphical means, by Professor 
Thurston. Comparing all these results and conclusions, it 
would seem, that, starting at the ordinary temperatures, — 

i°. A decrease of temperature increases the tensile strength 
of iron and steel, but decreases its ductility ; thus rendering it 
more brittle, and hence decreasing its resilience, or power of 
resisting shocks. 

2°. An increase of temperature up to about 570 F. in- 
creases the tensile strength ; but at a straw heat or a pale blue, 
almost all irons and steels are very brittle. 

3 . An increase of temperature above 576 decreases the 
tensile strength, but increases the ductility. 

Effect of Cold-Rolling on Iron and Steel. — It has already 
been stated, p. 370, that it was discovered independently by 
Commander Beardslee and Professor Thurston, that if a load 
were gradually applied to a piece of iron or steel which exceeded 
its elastic limit, and the piece then allowed to rest, the elastic 
limit and the ultimate strength would thus be increased. This 



STEEL. 



409 



may be accomplished with soft iron and steel by cold-rolling or 
cold-drawing, but cannot be taken advantage of in hard iron 
or steel. 

Professor Thurston, who has investigated this matter at 
great length, and made a large number of tests on the subject, 
gives the following as the results of cold-rolling : — - 



Increase in 


Per Cent. 




25 to 40 

50 to 80 

80 to 125 

300 to 400 

150 to 425 




Elastic limit (tension, torsion, and transverse), 


Elastic resilience (transverse) 



He also says, in regard to the modulus of elasticity, — 

" Collating the results of several hundred tests, the author 
[Professor Thurston] found that the modulus of elasticity rose, 
in cold-rolling, from about 25000000 lbs. per square inch to 
26000000, the tenacity from 52000 lbs. to nearly 70000, the 
elastic limit from 30000 lbs. to nearly 60000 lbs. ; and the ex- 
tension was reduced from 25 to \o\ per cent. 

" Transverse loads gave a reduction of the modulus of elas- 
ticity to the extent of about 1 000000 lbs. per square inch, an 
increase in the modulus of rupture from 73600 to 133600, and 
reduction of deflection at maximum load of about 25 per cent. 
The resistance of the elastic limit was doubled, and occurred 
at a much greater, deflection than with untreated iron." 

On the other hand, the two steel eye-bars referred to on 
p. 422 show a decrease of modulus of elasticity with increasing 
over-strain. 

Whitwortti s Compressed Steel. — Sir Joseph Whitworth pro- 
duces steel of great strength by applying to the molten metal, 
directly after it leaves the furnace, a pressure of about 14000 
lbs. per square inch ; this being sufficient to reduce the length 



4io 



APPLIED MECHANICS. 



of an eight-foot column by one foot. He claims, according to 
D. K. Clark, to be able to obtain with certainty a strength of 
40 English tons with 30 per cent ductility, and mild steel of a 
strength of 30 English tons with 33 or 34 per cent ductility. 

The following table is taken from D. K. Clark's " Rules 
and Tables : " — 





Ultimate Tensile 
Strength, in lbs., 
per Square Inch. 


Elongation, 1 
per cent. 

i 


Axles, boilers, connecting-rods, rivets, railway tires, \ 
guns, and gun-carriages, > 

Cylinder linings, parts of large machines, hoops, ) 
and trunnions, > 

Large planing and lathe tools, shears, smiths' "] 
punches, dies and sets, cold-chisels, screw tools, \ 
etc., J 

Boring-tools, finishing-tools for planing and turn- ) 
ing, ) 


89600 
IO7520 

I 2992O 

152320 
I 6l 280 


32 
24 

17 

IO 

14 





§226. Tensile Strength of Steel. — The older experiments 
on the strength of steel are of but little value ; as we have very 
imperfect records of the kind of steel used in the tests, and 
its mode of manufacture. Hence only the more recent experi- 
menters will be enumerated. 

Sir William Fairbairn : Useful Information for Engineers. 
David Kirkaldy : (a) Experiments on Wrought-Iron and Steel. 

(b) An Experimental Inquiry into the Strength of Fagersta Steel. 
W. E. Woodbridge : Report on the Mechanical Properties of Steel, 

Chiefly with Reference to Gun Construction on the Woodbridge 

System. 
Government. Machine : (a) Executive Document 23, 46th Congress, 2d 

session. 
(fi) Executive Document 5, 48th Congress, 1st session. 
J. Barba : The Use of Steel for Constructive Purposes. By J. Barba. 



TENSILE STRENGTH OF STEEL. 



411 



Tests made for St. Louis Arch : Woodward's History of the St. Louis 
Arch. 

Tests of Steel for the Brooklyn Bridge : Roebling's Report of the 
Brooklyn Bridge. 

Charles B. Dudley: Franklin Institute Journal, 1881. 

A. F. Hill : Proceedings of the Engineers' Society of Western Penn- 
sylvania, vol. i. 

Professor R. H. Thurston : Materials of Engineering. 

The following are the summaries of Kirkaldy's tests on 
steel bars and steel plates as given in his book, " Experiments 
on Wrought-Iron and Steel," published in 1866, giving the 
tensile strength and contraction of area of such specimens as 
he tested : — 

STEEL BARS. 



I 




eight per 
inch of 
irea. 


< 


eight per 
'nch of 
Area. 


a 
§ 


1 


Names of Makers or Works. 


*■ 2 


§ 2 


> " *a 


3 


i 




i 5 a = 

s § :& 

rt 0* J* 
£> CO O 

m 


■■5 8 

2 £ 

1 « 



bfl flJ 3 

•S C3 t! 
«J CO fa 

PQ 


a 
2 






lbs. 


%• 


lbs. 


%. 


6 


Turton & Sons, cast-steel for tools . 


132909 


47 


139124 


5-4 


4 


Jowitt, cast-steel for tools 


132402 


12.8 


151857 


5-2 


8 


Jowitt, cast-steel for chisels 








124852 


17.0 


150243 


7-i 


4 


Jowitt, cast-steel for drifts . 








1 1 5882 


21.5 


147570 


13-3 


4 


Jowitt, double shear steel . 








I 18468 


19.6 


147396 


13-5 


8 


Bessemer, Sheffield tool 








I I 1460 


22.3 


143327 


5-5 


4 


Wilkinson, blister steel . . 








104293 


21.4 


132472 


9-7 


4 


Jowitt, cast-steel for taps . 








101151 


28.8 


142070 


10.8 


4 


Krupp's cast-steel for bolts 








92015 


34-o 


139434 


15-3 


4 


Shortridge & Co.'s homogeneous metal, 


90647 


36.6 


142920 


J 3-7 


4 


u u u u 


89724 


26.0 


121212 


1 1.9 


4 


Jowitt's spring steel 


72529 


24.1 


95490 


18.0 


6 


Mersey Company's puddled steel . . 


71486 


35-3 


110451 


19.1 


6 


Blochairn puddled steel 


70166 


19.4 


84871 


"•3 


6 


<« « « 


65255 


19.0 


80370 


12.0 


4 


<« « « 


62769 


1 1.9 


71231 


9.1 



412 



APPLIED MECHANICS. 



STEEL PLATES. 



Names of Makers or Works. 



Turton & Son, cast-steel . . 
Shortridge & Co., cast-steel . 
Naylor, Vicker, & Co., cast-steel 
Moss & Gambles, cast-steel 
Shortridge & Co., cast-steel . 
Mersey Company, puddled steel 
Mersey Company, hard steel . 
Blochairn, hard steel .... 



Shortridge & Co., hard steel 
Mersey Company, mild . . 



£"!5 



« in O 
pq 



95299 
96715 
84435 
72338 
96989 
93209 

93979 
933*6 

85010 

72994 
72366 

7i53 2 



< 

.2 o 

I « 
u 



%. 

9-5 
15.2 
21.7 

334 
14.4 

54 
4.6 
4.8 

7-i 
8.6 
10.5 

7-5 



£ C/3 fa 

PQ 



lbs. 

105937 
I 14203 

108125 
109050 

"3395 

100649 
98472 
97978 
92130 
80034 
80937 
7775° 



fa 



%. 

7.68 

8-77 
17.41 

19-73 

14.40 

2.02 

4.08 

3-14 
6.18 

4-57 
5-94 
3-57 



These tests were made before the use of steel in construc- 
tion had become as general as it has at the present day. 

The quality and strength of steel are affected very seri- 
ously by its chemical composition. In this regard we have 
the record of a series of tests made by the Committee on 
Chemical Research of the United-States Government Com- 
mission, and recorded in Executive Document 23, 46th Con- 
gress, second session. 

The following summary of this report will be appended here, 
giving only the percentages of the most important ingredients 
besides the iron. This summary is as follows : — 



TENSILE STRENGTH OF STEEL. 



413 





u 

a | 

v c 

> ■- 
< 


"8 

Ji 

"3 

-a 
. 

i) M 

2 rt 


tics 


S « 

•>C/2 O 


Chemical Analysis. 


I"! 




•3 

s 

O 


e 

J 


3 

% 

3 

C/5 


3 

1 
1 


a 

rt 












% 


% 


% 


% 


% 


% 


1518 


21600 


24648000 


37933 


96033 


0.246 


O.OII 


0.145 


O.OO4 


0.014 


0.020 


1519 


40000 


28010000 


67666 


109166 


1 0.383 
1 0.389 


1 0.006 


0.076 


O.OO3 


0.017 


0.060 


1520 


37052 


26578000 


56659 


123433 


( 0.276 

J 0.277 


1 0.018 


0.061 


O.OOI 


0.015 


0.031 


1521 


44577 


26750000 


67817 


139633 


i o-377 
1 0.372 


£ O.O25 


0.070 


0.002 


0.014 


0.027 


1522 


35854 


25799000 


60214 


1 21 800 


S °- 2 79 
/ 0.273 


> 0.022 


0.070 


0.002 


0.015 


0.041 


1523 


41721 


26507000 


78119 


1 1 4300 


| 0.468 
| 0.464 


f O.O27 


0.090 


O.OOI 


0.015 


0.032 


1524 


45090 


26013000 


83497 


129933 


\ 0.728 
1 0.724 


r 0.028 


0.126 


0.002 


0.014 


0.048 


1525 


49544 


24816000 


90940 


122133 


I0.611 

| 0.608 


| 0.027 


010 


0.003 


0.014 


0.042 


I526 


51282 


24588000 


95340 


130600 


(0676 
} 0.678 


J 0.018 


0.135 


0003 


0.015 


0.038 


1527 


51609 


27049000 


100719 


121900 


| 0.557 
f 0.551 


\ O.027 


0.112 


0.002 


0.015 


0.028 


1528 


51076 


26587000 


115169 


138133 


I 0.833 
| 0.828 


f O.040 


0.134 


Trace 


015 


0.046 


1529 


523" 


25979000 


120275 


141433 


i 0.908 
| 0.909 


j 0.015 


0.1 41 


0.002 


0.014 


0.036 


I530 


48597 


26648000 


123697 


144492 


0.924 


0.022 


0.187 


0.002 


0.014 


0.031 


1531 


55954 


24682000 


118156 


140367 


0.966 


O.030 


0.153 


Trace 


0.015 


0.035 


1532 


54596 


26819000 


117660 


141867 


0.946 


0.930 


0.162 


O.OOI 


0.015 


0.029 


1533 


60627 


26250000 


1 1 6636 


138600 


1.024 


O.020 


0.173 


Trace 


0.013 


0.014 


1534 


59160 


25576000 


110823 


118500 


1.079 


0.027 


0.096 


O.OOI 


0.014 


0.044 


1535 


66710 


27091000 


120602 


132060 


1. 112 


O.030 


0.190 


Trace 


015 


0.045 


1536 


63885 


25186000 


119514 


128800 


1. 186 


O.024 


0.114 


Trace 


6.014 


0.132 


1537 


70405 


25824000 


121253 


128800 


1.285 


O.033 


0.106 


Trace 


0.015 


0.273 


io53 


58971 


28169000 


103761 


149400 




f o-973 ) 




0.213 


0.003 


0.025 


0.073 


1054 


59040 


28608000 


106205 


167367 




0886 




0.196 


0.002 


0.037 


0.185 


1055 


51500 


31368000 


92081 


151467 


a 




0.694 




128 


Trace 


0.037 


0.137 


1056 


57802 


27668000 


94529 


137900 


,g 


o.994 




0.140 


0.003 


0.027 


O.IOI 


1057 


50494 


27776000 


72979 


151433 


6 • 


0.401 




0.085 


0.006 ' 


0.032 


O.II2 


1058 


54763 


26574000 


96373 


152233 


1 


0.905 




0.161 


Trace 


0.026 


O.I08 


io59 


54243 


29028000 


95612 


127133 


H 


0.915 




0.191 


0.002 


0.026 


O.086 


1069 


50286 


25538000 


70066 


102967 




0.238 




0.105 


0.012 


0.034 


O.184 


106 1 


40667 


24536000 


67000 


120067 




L 0.463 J 




0.121 


0.002 


0.020 


Trace 


1065 


35167 


26379000 


55000 


140600 


0.184 


0.009 


0.063 


Trace 


0.014 


0.051 


1066 


48367 


25637000 


83660 


139267 


0.459 


0.118 


0.108 


Trace 


0.026 


0.185 


1067 


50583 


28546000 


81216 


150367 


0451 


O.003 


0.134 


0.004 


0.026 


0.139 



414 



APPLIED MECHANICS. 



O CO 
H-l 


J J3 

v s 

> •- 


"3 

O . 

« .5 

SJ 3 



• - . ri 

■£•§. 

V MCE 

So c^ 
2 £ 
£co"o 
< 


.S u 

w ah 
& e >- 
2co"o 


Chemical Analysis. 


u c 

H 

1<3- 


I § 

2U 



a 
| 

CO 


u 

3 

3 
CO 


.3 
c 

ft 

O 




e 

n 

So 
a 

rt 

3 












% 


% 


% 


% 


% 


% 


1068 


50522 


28755000 


111910 


137233 


o.793 


0.013 


0.172 


None 


0.019 


0.193 


1069 


5M35 


25627000 


1 1 7668 


143633 


0.825 


0.009 


0.170 


None 


0.020 


0.182 


1070 


55*37 


26498000 


126577 


I5H33 


0.846 


007 


0233 


None 


0.019 


0.191 


1071 


62691 


28219000 


129837 


151800 


o.959 


0.013 


0.225 


Trace 


018 


0.213 


1072 


61062 


27643000 


121416 


148900 


984 


0.008 


0.157 


None 


0.019 


0.245 


1073 


68015 


26805000 


121920 


132000 


1.059 


0.013 


0.162 


None 


0.022 


0.252 


1074 


69427 


27552000 


126036 


144300 


1. 108 


0.013 


0.206 


None 


0.023 


0.269 


io75 


72700 


27760000 


129293 


145567 


1. 142 


0.012 


0.204 


None ' 


0.020 


0.282 


1076 


75403 


27264000 


132010 


138600 


1.244 


0.082 


0.246 


None 


0017 


0.262 


1077 


50000 


28102000 


96699 


101800 


0.627 


0.012 


0.154 


Trace 


0.007 


0.050 


1078 


49436 


28701000 


9^38 


169733 


o-439 


0.015 


0.136 


0.001 


0.019 


0.023 


1079 


36941 


27525000 


80945 


135167 


0.461 


None 


0.1 16 


Trace 


0.020 


0.027 


1080 


25167 


25953000 


43417 


88500 


^ 0.209 
I car 


= total ) 
x>n. ) 


0.163 


0.009 


0.084 


0.020 


1081 


28333 


29028000 


51833 


149200 


0.116 


0.014 


O.OII 


0.029 


0045 


0.192 


1082 


51609 


28518000 


102240 


109333 


0.675 


0.016 


0.028 


0.028 


0.065 


0.459 


1083 


61388 


26295000 


128750 


139833 


670 


O.OII 


0.043 


0.014 


0.063 


0.625 


1084 


34333 


24589000 


54333 


99800 


0.049 


0.008 


0.219 


0.007 


0.179 


0.063 


1085 


48350 


28752000 


85833 


155267 


o.433 


0.014 


071 


0.008 


0.062 


0.493 


1086 


44667 


25360000 


65000 


131833 


O.III 


0.012 


0.006 


0.031 


0.123 


103 


1087 


54217 


25400000 


88007 


92967 


0.348 


0.007 ' 


0.028 


0.041 


0.125 


0.404 


1088 


65190 


23648000 


101045 


102300 


0.744 


O.OI2 


0.074 


0.043 


0.104 


.0.465 


1089 


27617 


23341000 


48082 


97033 


0.042 


None 


0.168 


O.OII 


0.109 


0.051 


1090 


30000 


24071000 


51500 


75833 


0.012 


0.003 


0.214 


0.048 


0.315 


0.081 


1091 


26000 


25482000 


59000 


80033 


0.230 


0.004 


0.084 


Trace 


0.039 


None 


1092 


54869 


25503000 


79315 


84733 


0.314 


0.012 


0.016 


0.066 


0.099 


0.525 


1093 


48733 


29326000 


78500 


92167 


0.243 


O.OII 


0.013 


0.058 


0.128 


0.34* 


1094 


47833 


26110000 


68667 


88467 


o.i35 


0.004 


0.007 


0.056 


0.113 


0.165 


io95 


56505 


25917000 


95341 


101633 


o.374 


008 


018 


0.062 


0.138 


0.584 


1096 


63126 


24602000 


111203 


142133 


o.375 


0.012 


0.070 


0.038 


0.092 


0.685 


1582 


37919 


28697000 


55955 


182533 


- 


- 


- 


- 


- 


- 


1583 


33247 


28205000 


50523 


161700 


- 


- 


- 


- 


- 


- 


1588 


60627 


27010000 


110389 


162900 


- 


- 


- 


- 


- 


- 


1589 


58562 


25790000 


106695 


176467 


- 


- 


- 


- 


- 


! 


1590 


50305 


25772000 


79858 


154267 


- 


- 


- 


- 


- 




1591 


60627 


27389000 


123645 


142533 


- 


- 


- 


- 


- 


- 


i59 2 


58345 


25323000 


103761 


147167 


- 


- 


- 


- 


- 


- 


1593 


30933 


24633000 


49167 


95600 


- 


- 


- 


- 


- 


- 


1594 


53565 


25776000 


78608 


148300 


- 


- 


- 


- 


- 


- 


1595 


46394 


25330000 


64104 


129200 


" 


" 


" 


" 







TENSILE STRENGTH OF STEEL. 



415 



Mr. A. F. Hill, in the "Transactions of the Engineers' 
Society of Western Pennsylvania" for 1880, gives an account 
of some tests of open-hearth steel manufactured by Anderson 
& Co. of Pittsburgh, taken from different runs, and ranging 
from 0.3 to 0.5 per cent of carbon ; the steel being tested both 
in the form of specimens, and also in the form of eye-bars, 
plates, and riveted plate-girders. 

The following account of some of these tests is condensed 
from his paper : — 



I. SPECIMEN TESTS OF BAR-ENDS 30 INCHES LONG, DESIGNED FOR 

EYE-BARS. 



Mark, and Carbon 


Tensile Stress per Square Inch 
at 


Stretch at 


Reduction 
of Area at 


Percentage. 


Elastic Limit. 


Rupture. 


Fracture, %. 


Fracture, %. 


I 




55712 


94760 


i5-i 


30 


2 




56009 


95380 


12.9 


26 


3 


\ 0.3% c. 


55120 


95830 


1 5-3 


31 


4 




55830 


96020 


14.5 


27 


5 J 




55512 


94970 


13.8 


29 


1 




6579O 


1 1 2340 


10.8 


19 


2 




6604O 


1 1 2470 


8.9 


16 


3 


• 0.5% c. 


66160 


1 1 1980 


10.5 


22 


4 




65550 


I I3320 


10.9 


21 


5 J 




65980 


I I 3040 


9.4 


20 



416 



APPLIED MECHANICS. 



II. EXPERIMENTS ON STEEL EYE-BARS WITH 0.3% CARBON. 

Dimensions : Stem, 3 in. X f in. X 10 ft. Head, \\ in. thick, 7^ in. across the eye. 
Pin-hole, 3! in. diameter. 



Mark. 


3 
C 

3 


Tensile Stress per 
Square Inch at 




2 

£ 3 
CO 


J* 

O £ 
3 H. 

.2 £ 

3 *** 
(3 


Remarks. 


Elastic 
Limit. 


Rupture. 


Bl 
B2 
B3 


1/1 


54026 
54H3 
54II3 


87400 
94500 
89300 


8.2 
9.2 
7.0 


44 
46 
42 


Broke 1 ft. 3 in. from pin-hole. 
- 4ft. 3 tin. " 
" 1 ft. 8± in. " " 


Fl 
F2 
F3 


'o 


51762 
54065 
52518 


92672 
91570 
94780 


8.2 

9-3 
1 1.8 



29 

40 


Broke in head into 3 pieces. 
" 5 ft. z\ in. from pin-hole. 
" 2 ft. if in. " 


Al 
A2 
A3 




58473 
56059 

553™ 


69140 
63OOO 
69400 


2.0 
2.6 
2.2 


<u O 
»i c *s 

So to « 

<u « rt 

s 1 ** 


Broke 5^ in. from pin-hole. 
" - 5 lin. " 
« 5 |in. « 



In those marked "upset,'' the bars were rolled to the re- 
quired section of the stem, with sufficient surplus of length 
to form the heads by hydraulic upsetting. 

In those marked " rolled," the heads were rolled by Klo- 
man's patent process. 

In those marked " welded," the heads were formed by weld- 
ing pieces, and die forging. 

Fractures of Bi, B2, B3, ¥2, and F3 were fine, silky, and 
wedge-shaped. 

Fi broke in head, and showed effect of over-heating; frac- 
ture coarse and granular. 

A 1, A2, and A3 broke in stem close to neck ; fracture close- 
grained, 50 per cent granular, and showing effect of welding 



TENSILE STRENGTH OF STEEL. 417 

heat Weld was defective at junction of head and neck on 
account of welding-pieces having been too small. 

Mr. Hill draws from these tests the following conclusions : — 

i°. "The strength of the specimen exceeds, in each case, 
that of the manufactured bar." 

2°. "The uniformity of the results obtained from the tests 
of the bar-ends shows conclusively that whatever difference in 
strength there is between these bar-ends and the manufactured 
eye-bar is properly ascribable to the mode of manufacture." 

3 . " The results obtained from the rolled and the upset eye- 
bars approach nearest to the original bar strength, and give the 
best results. The difference between the results from these two 
methods is so trifling — and if any thing in the 0.3 per cent car- 
bon group, slightly in favor of the upset bar — that it leaves no 
doubt in my mind that these two processes are equally good." 

4 . " The results from the welded bars show, that, while 
steel can be perfectly welded, there is a loss of nearly 30 per 
cent of ultimate strength as compared with the original bar; 
moreover, the elastic limit is too near the ultimate strength, and 
the percentage of elongation too small, to give sufficient warn- 
ing of impending failure." 

" It will, therefore, be safe to conclude that welded members 
in steel construction, while no worse than welded iron ones, are 
not desirable, and, in fact, ought not to be admitted at all, 
except where the grade of steel used is very low ; and then the 
greatest caution in working and annealing will be required." 

Mr. Hill states also that his experiments with eye-bars con- 
taining 0.5 per cent carbon bear out these conclusions. 

Tests of Steel Eye-Bars made on the Government Machine. — 
In Executive Document No. 5, 48th Congress, first session, is 
the record of the tests of six eye-bars of steel, presented by 
the president of the Keystone Bridge Company. 

The following is an extract from the report in regard to 
these eye-bars : — 



4i8 



APPLIED MECHANICS. 



"The eye-bars were made of Pernot open-hearth steel, fur- 
nished by the Cambria Iron Company of Johnstown, Penn. 

"The furnace charges, about .15 tons each of cast-iron, 
magnetic ore, spiegeleisen, and rail-ends, preheated in an aux- 
iliary furnace, required six and one-half hours for conversion. 

"All these bars were rolled from the same ingot. 

" Samples were tested at the steel-works taken from a test 
ingot about one inch square, from which were rolled |-inch 
round specimens. 

"The annealed specimen was buried in hot ashes while still 
red-hot, and allowed to cool with them. 

"The following results were obtained by tensile tests : — 



f-inch round rolled bar . 

f-inch round rolled and 

annealed bar .... 



Elastic 
Limit, in 
lbs., per 

Sq. In. 



4804O 
42210 



Ultimate 

Strength, in 

lbs., per 

Sq. In. 



73150 
69470 



Contrac- 
tion of 
Area. 



457 
54-2 



Modulus 

of 
Elasticity. 



282IOO0O 
292IOOOO 



Carbon. 



O.27 
O.27 



"The billets measured 7 inches by 8 inches, and were 
bloomed down from 14-inch square ingot. 

"They were rolled down to bar-section in grooved rolls at 
the Union Iron Mills, Pittsburgh. 

" The reduction in the roughing-rolls was from 7 inches by 
8 inches to 6\ inches by 4 inches ; and in the finishing-rolls, to 
6J inches by 1 inch. 

"The eye-bar heads were made by the Keystone Bridge 
Company, Pittsburgh, by upsetting and hammering, proceeding 
as follows : — 

"The bar is heated bright red for a length of (approxi- 
mately) 27 inches, and upset in a hydraulic machine ; after 



TENSILE STRENGTH OF STEEL. 419 

which the bar is reheated, and drawn down to the required 
thickness, and given its proper form in a hammer-die. 

"The bars are next annealed, which is done in a gas-furnace 
longer than the bars. They are placed on edge on a car in the 
annealing-furnace, separated one from another to allow free 
circulation of the heated gases. They are heated to a red heat, 
when the fires are drawn, and the furnace allowed to cool. 
Three or four days, according to conditions, are required before 
the bars are withdrawn. 

"The pin-holes are then- bored 

"The analyses of the heads before annealing were: — 

" Carbon, by color 0.270 per cent 

Silicon 0.036 " 

Sulphur °-°75 u 

Phosphorus . 0.090 " 

Manganese 0.380 " 

Copper Trace. 

" The bars were tested in a horizontal position, secured at 
the ends, which were vertical. 

" To prevent sagging of the stem, a counterweight was used 
at the middle of the bar. 

" Before placing in the testing-machine, the stem from neck 
to neck was laid off into 10-inch sections, to determine the 
uniformity of the stretch after the bar had been fractured. 

" A number of intermediate 10-inch sections were used as 
the gauged length, obtaining micrometer measurements of 
elongation, and the elastic limit for that part of the stem which 
was not acted upon during the formation of the heads. Elon- 
gations were also measured from centre to centre of pins, taken 
with an ordinary graduated steel scale. 

"The moduli of elasticity were computed from elongations 
taken between loads of 10000 and 30000 lbs. per square inch, 
deducting the permanent sets. 



42 O APPLIED MECHANICS. 

"The behavior of bars Nos. 4582 and 4583, after having 
been strained beyond the elastic limit, is shown by elongations 
of the gauged length measured after loads of 40000 and 50000 
lbs. per square inch had been applied; and with bar No. 4583, 
after its first fracture under 64000 lbs. per square inch, a rest 
of 'five days intervening between the time of fracture and the 
time of measuring the elongations. 

"Considering the behavior between loads of 10000 and 
30000 lbs. per square inch, we observe the elongations for the 
primitive readings are nearly in exact proportion to the incre- 
ments of load. 

" Loads were increased to 40000 lbs. per square inch, passing 
the elastic limit at about 37000 lbs. per square inch ; the respec- 
tive permanent stretch of the bars being 1.3 1 and 1.26 per cent. 

" Elongations were then immediately redetermined, which 
show a reduction in the modulus of elasticity, as we advanced 
with each increment, of 5000 lbs. per square inch. 

"Corresponding measurements after the bars had been 
loaded with 50000 lbs. per square inch reach the same kind of 
results. 

"The first fracture of bar No. 4583, under 64000 lbs. per 
square inch, occurred at the neck, leaving sufficient length to 
grasp in the hydraulic jaws of the testing-machine, and con- 
tinue observations on the original gauged length. This was 
done after the fractured bar had rested five days. 

"The elongations now show the modulus of elasticity con- 
stant or nearly so, the only difference in measurements being 
in the last figures, up to 50000 lbs. The readings were then 
immediately repeated, and the same uniformity of elongations 
obtained. 

" An illustration of the serious influence of defective metal 
in the heads is found in the first fracture of bar No. 4583. 

" There was about 27 per cent excess of metal along the 
line of fracture over the section of the stem." 



TENSILE STRENGTH OF STEEL. 



421 









-^ 


4>- 


+. 


+. 


■*>■ 


H ? 




00 


CO 


CO 


00 




ob 


cn 


en 

CO 


$ ° 3 


"*' 


CN 


en 




JCO 
nV, 






N 


r "*" a* 


to 


Kl 


N) 


K) 






K) 


M 


3" r 


ON 






ON 










-. 8 


VI 


to 


00 


V) 



' 


1 


i? 


00 
00 


^ g. 


8 


8 


to 

8 


8 


1 


1 


On 
O 


ON 




Gauged Length, in inches. 


ON 


on 


0* 


ON 






On 


ON 


Width, in inches. 


o\ 


Ul 


On 


ON. 






On 


00 










O 


O 






O 





I -. g ^ 






NO 












S^ 3 ffi 0' 


VI 


vl 


ON 


00 






CO 


00 


8 >? w 


VI 

on 

■0 


to 

"8 


00 




vl 

\ 


1 


1 


ON 

*0 


CO 

0* 


_ co 5' r h 

s *2 "3 « 5- ST 

£• § 3 5 s 3. w 

3 ; » 0" 


















w s-~ ^ 


















a* ■§ ^ - 1 !■ 


00 


* 






to 






vl 



























D 


en 

O 


8^ 


8 


8 


' 3 -* S-ff 


w 


ON 


to 


to 




w 




Cn 


In Gauged 


« S 


■8 


O 










1 


v) 


1 


S 5 


Length. 


• 5- S" 


CO 


en 


« 


10 








Cn 


Centre to Cen- 


Cn 


H 


en 


CT> 








*o 


tre of Pins. 


i 3 


to 




CO 


to 


(00 


ON 




CO 


Contraction of Area, per 


*- 


0, 


» 


M 


*■ 


Or 




vl 


cent. 


vo 


•O 


NO 


£ 








g 


ST £ 


0. 


vl 

8 
8 


ON 




1 


I 


| 


00 


^ EL 


00 


v] 














Maximum Compression 


10 







On. 


1 


1 


NO 


ON 


on Pin»Holes, in lbs., 


2 








Q 






CO 
O 


8 


per Square Inch. 












u 


bd 


w 

















a 


3 
















IT 


ST 


M *^ 














w 


















3 


3 


8 1: 












» 


d 


VI 


c § 












3 


3 

O 
7? 


ft 

3 


a s 
P 


s 


51 


Tj 


CO 


*j 


^ 


O 


CO 


> 


m 


3 
n 


3 

n 




e 3 
ET « 


3 



W 

3 


>< 


a 


?? 


a 


IB 

7? 


65 

3 


6' h 

? 5? 




ST 


3 


| 


" 3 




3 


OP 


I 


3 




Crq 


8 






Cl 


c 




0. 




c 




Orq 




W 






ft 






•n 


2 














M 


3 


s 




3 




OP 











c 












































3 


p 












* 




c 


•* 











422 



APPLIED MECHANICS. 



ELONGATIONS OF No. 4582 FOR EACH INCREMENT OF 5000 LBS. PER 

SQUARE INCH. 



Loads, in lbs., 

per 
Square Inch. 


Elongations. 


Primitive Load- 
ing. 


After Load of 
40000 lbs. per 
Square Inch. 


After Load of 
50000 lbs. per 
Square Inch. 


I OOOO 
150OO 
2O0O0 
250OO 
3000O 


O.0274 
O.0269 
O.0269 
O.0269 


O.0300 
O.O305 
O.0320 
O.0330 


O.O31 1 
O.0322 
O.0337 
O.034I 



ELONGATIONS OF No. 4583 FOR EACH INCREMENT OF 5000 LBS. PER 

SQUARE INCH. 



Loads, in 

lbs., per 

Square Inch. 


Elongations. 


Elongations after 64000 lbs. per 
Square Inch. 


Primitive Load- 
ing. 


After 40000 lbs. 

per 

Square Inch. 


After 50000 lbs. 

per 

Square Inch. 


First 
Reading. 


Second 
Reading. 


I OOOO 
15OOO 
20000 
25000 
30000 
350OO 
4OOOO 
4500O 
50O00 


O.0272 
O.0272 
O.0268 
O.O267 


O.029I 
O.0305 
O.O314 
O.O326 


O.O302 
O.O315 
O.O325 
O.0340 


O.O3 1 1 
O.0308 
O.03 1 1 
O.O312 
O.031 1 
O.0312 
O.03IO 
O.O315 


O O p 
b b b b 

0000 



STEEL PLATES. 



Steel plates are used in making plate-girders and other 
forms for resisting load, and also for steam-boilers. 



STEEL PLATES. 



423 



For the latter purpose the steel must be very low in carbon, 
and must stand very much more in the way of bending, punch- 
ing, shearing, etc., than in the former case. Its tensile strength 
will not be high, but it must have great ductility. The first set 
of tests to which reference will be made is the set given in Mr. 
Hill's paper already referred to, this steel being too high to be 
suitable for boiler-work. 

He had made 54 tests of rolled open-hearth steel plates. 

The following 15 tests give the relative strengths of the 
plates, and the percentages of carbon. 

The plates tested were all | inch thick, 12 inches wide, and 
6 feet long, tested as they came, from the rolls; crop ends 
sheared, 50 inches between jaws of machine. 



Mark. 


Carbon, 
percentage. 


Tensile Stress, in lbs., per 

Square Inch, in Direction 

of Rolling, at 


Average, per 

cent, 

Elongation at 

Fracture. 


Remarks. 


Elastic Limit. 


Rupture. 


Pi 
P2 
P3 
P4 
P5 
Ri 
R2 

R3 
R 4 

R5 
Vi 
V2 
V3 
V 4 
V5 


0.3 

o-3 
0.3 
o-3 
0.3 
0.4 
0.4 
0.4 
0.4 
0.4 
0.5 
0.5 

0:5 

0.5 

o-5 


43260 
44820 
45UO 
43990 
44720 
51620 
50980 
51260 
51 100 
5089O 

58950 
59200 
58540 
58880 
59330 


79120 
77840 
78390 
77970 
78280 
81990 
81720 
83730 
81830 
83130 
85790 
86220 
85560 
86000 
86330 


' 19-3 
' 13-9 
- 10.5 


Fractures fine and silky. 

Fractures very fine. 

(Fractures good, slightly 
I granular on edges. 



424 



APPLIED MECHANICS. 



He then gives the following table : — 



COMPARATIVE RESULTS OF SHEARING, PUNCHING, ANNEALING, AND 
TEMPERING STEEL PLATE. 



Carbon, 
per cent. 



Treatment of Specimen. 



Tensile Stress per Square 
Inch at 



Elastic Limit 



Rupture. 



o-3 
o-3 
o-3 
o-3 
°-3 
0.4 
0.4 
0.4 
0.4 
0.4 
o-5 
o-5 
°-5 
°-5 
°-5 



Cut in planer 

Sheared 

Punched 

Punched and hammered cold . . 
Punched, hammered, and annealed, 

Cut in planer 

Sheared 

Punched ... 

Sheared and annealed . . . . . 

Punched and tempered 

Cut in planer 

Sheared 

Punched 

Sheared and tempered 

Punched and annealed 



4943 1 

3 2 37o 

o 

o 

5578o 

63475 

46900 

o 

59350 
52780 
65T86 
51666 

o 

60375 
57960 



94396 
74980 
63410 
87540 

100410 

87695 

75330 
68890 

86160 

103560 



79COO 
78400 

87293 

84900 



From these tests he draws the following conclusions : — 

i°. "That both shearing and punching are injurious to all 
grades of steel, and cold punching far more than shearing." 

2°. " That both these operations affect the elastic limit far 
more than they do the ultimate strength." 

3 . "That apparently the lower grades of steel are more in- 
jured than higher grades ; but the evidence on this point is not 
certain, as the lower-grade plates were thicker than those of 
higher grade." 



FACTOR OF SAFETY. 425 



TRANSVERSE STRENGTH OF STEEL. 

A number of tests of the transverse strength of small bars 
of steel 1 or 2 inches square have been made by Kirkaldy and 
others ; and they have generally shown a modulus of rupture 
larger than the tensile strength of the steel, frequently 1.5 
times the tensile strength. 

It is only now that the attempt is being made to roll steel 
I-beams, and thus far no tests have been made on full-size steel 
beams. Before we can decide upon the value of the modulus 
of rupture suitable to use, we need such tests ; and it would not 
be safe to assume for these beams the constants deduced from 
the small ones. The steel used may vary greatly from a very 
mild to a very hard steel. It is probable that it will be a mild 
steel that will be used for beams. 

Until we have experiments upon full-size beams, it would 
hardly be safe to use, for modulus of rupture, much more than 
would hold for an iron beam, or from 50000 to 60000 or 70000 
lbs. per square inch. 

§ 227. Factor of Safety. — In order to determine the 
proper dimensions of any loaded piece, it becomes necessary 
to fix, in some way, upon the greatest allowable stress per 
square inch to which the piece shall be subjected. 

The most common practice has been to make this some 
fraction of the breaking-strength of the material per square 
inch. 

As to how great this factor should be, depends upon — 

i°. The use to which the piece is to be subjected ; 

2°. The liability to variation in the quality of the material ; 

3 . The question whether we are considering, as the load 
upon the piece, the average load, or the greatest load that can 
by any possibility come upon it ; 

4 . The question as to whether the structure is a temporary 
or a permanent one ; 



426 APPLIED MECHANICS. 

5°. The amount of injury that would be done by breakage 
of the piece ; 
and other considerations. 

The factors most commonly recommended are, 3 for a dead 
or quiescent load, and 6 for a live or moving load. 

The American and English practice in the case of iron 
bridges is to use a factor of safety of 4 for both dead and 
moving load. In machinery a factor as large as 6 is desirable 
when there is no liability to shocks ; and when there is, a larger 
factor should be used. 

A method very rarely followed for tension and compression 
pieces is, to prescribe that the stretch under the given load 
should not exceed a certain fixed fraction of the length. This 
requires a knowledge of the modulus of elasticity of the mate- 
rial. 

In the case of a piece subjected to a transverse load, it is 
the most common custom to determine its dimensions in accord- 
ance with the principle of providing sufficient strength; and 
for this purpose a certain fraction (as one-fourth) of the mod- 
ulus of rupture is prescribed as the greatest allowable safe 
stress per square inch at the outside fibre. Thus, the two iron 
companies already referred to prescribe 12000 lbs. per square 
inch as the greatest allowable stress at the outside fibre for 
wrought-iron beams, this being about one-fourth of the modulus 
of rupture. 

The other method for dimensioning a beam is, to prescribe 
its stiffness ; i.e., that it shall not deflect under its load more 
than a certain fraction of the span. This fraction is taken as 

4 t0 6 0* 

This latter method depends upon the modulus of elasticity 
of the beam ; and while it is the most advisable method to 
follow, and as a rule would be safer than the other method, 
nevertheless, in the case of very stiff and brittle material it 
might be dangerous : hence we ought to know also the break- 



WOHLER'S RESULTS. 427 

ing-weight and the limit of elasticity of the beam we are to use, 
and not allow it to approach either of these. This precaution 
will be especially important to observe in the case of steel 
beams, which are only now being introduced. 

On the other hand, in moving machinery a factor of safety 
of six is usually required when there is no unusual exposure to 
shocks, as in smooth-running shafting, etc. ; and when there 
are irregular shocks liable to come upon the piece, a greater 
factor is used. 

§228. Wohler's Results. — The extensive experiments of 
Wohler for the Prussian government, which were subsequently 
carried on by his successor, Spangenberg, were made to deter- 
mine the effect of oft-repeated stresses, and of changes of 
stress, upon wrought-iron and steel. 

In the ordinary American and English practice, it is cus- 
tomary, in determining the dimensions of a piece, as of a bridge 
member, to ascertain the greatest load which the piece can 
ever be called upon to bear, and to fix the size of the piece in 
accordance with this greatest load. 

Wohler called attention to the fact that the load that would 
break a piece depends upon both the greatest and least load 
that it would ever be called upon to bear. Thus, a tension-rod 
which is subjected to alternate changes of load extending from 
20000 to 80000 lbs. would require a greater area for safety than 
one which was subjected to loads varying only between the 
limits of 60000 and 80000 lbs. ; and this would require more 
area than one which was subjected to a steady load of 80000 
lbs. 

Wohler expresses this law as follows, in his " Festigkeits 
versuche mit Eisen und Stahl." 

"The law discovered by me, whose universal application 
for iron and steel has been proved by these experiments, is as 
follows : The fracture of the material can be effected by 
variations of stress repeated a great number of times, of 



428 APPLIED MECHANICS. 

which none reaches the breaking-limit. The differences of 
the stresses which limit the variations of stress determine the 
breaking-strength. The absolute magnitude of the limiting 
stresses is only so far of influence as, with an increasing stress, 
the differences which bring about fracture grow less. 

"For cases where the fibre passes from tension to compres- 
sion and vice versa, we consider tensile strength as positive 
and compressive strength as negative ; so that in this case the 
difference of the extreme fibre stresses is equal to the greatest 
tension plus the greatest compression." 

Besides the ordinary tests of tensile, compressive, shearing, 
and torsional strength, he made his experiments mainly on the 
following two cases : — 

i°. Repeated tensile strength; the load being applied and 
wholly removed successively, and the number of repetitions 
required for fracture counted. 

2°. Alternate tension and compression of equal amounts 
successively applied, the number of repetitions required for 
fracture being counted. 

In making these two sets of tests, he made the first set in 
two ways : — 

(a) By applying direct tension. 

(b) By applying a transverse load, and determining the 
greatest fibre stress. 

The second set of tests was made by loading at one end a 
piece of shaft fixed in direction at the other, and then causing 
it to revolve rapidly, each fibre passing alternately from tension 
to an equal compression, and vice versa. 

He also tried a few experiments where the lower limit of 
stress was neither zero nor equal to the upper limit, with a 
minus sign, also some experiments on torsion, on shearing, 
and on repeated torsion. 

When Wohler had made his experiments, and published his 
results, there were a number of attempts made by different 



LAUNHARDT'S FORMULA. 429 

persons to deduce formulae which should depend upon these 
experiments for their constants, and which should serve to 
determine the breaking-strength for any given variation of 
stresses. 

Only two of these formulae will be given here, viz., — 
i°. That of Launhardt for one kind of stress, 
2°. That of Weyrauch for alternate tension and compression, 
as these are the most used of the formulas developed. 



LAUNHARDT'S FORMULA. 

The constants used in this formula are — 

i°. t, the carrying-strength of the material per unit of area ; 
this being the greatest load per square inch of which the piece 
can bear an unlimited application without breaking. Practi- 
cally it is the breaking-strength per unit of area. 

2°. ?/, the primitive safe strength ; i.e., the greatest stress 
per unit of area of which the piece can bear, without breaking, 
an unlimited number of repetitions, the load being entirely 
removed between times. It is not a safe but a breaking 
strength. These two quantities have been determined experi- 
mentally by Wbhler; and it is the object of Launhardt's formula 
to deduce, in terms of t, u, and the ratio between the greatest 
and least loads to which the piece is ever subjected, the value 
a of the breaking-strength per unit of area when these loads 
are applied. 

The formula and its deduction are as follows : — 

Let the greatest stress per unit area be a. 
the least stress per unit area be c. 
their difference be d = a — c. 

Now a depends on d for its value ; and hence we may write 

a = ad, 
where a is a function of a also. 



43 O APPLIED MECHANICS. 



Now, the two 


cases 


experimented 


upon, 


viz., 


— 




i°. When a = 


t, 


c= t, 






</== 


o, 


2°. When a = 


u, 


c = o, 






</ = 


u, 



must be satisfied by the value of a used : otherwise the value 
we use is wrong, as these are two particular cases. Now, Laun- 
hardt takes 





= 


t 


— 


u 




t 


— 


a' 


a 




t 


— 


u 
-d 



and hence 

t — u 

t-a- « 

and it will become evident that this value of a does satisfy the 
two conditions stated if we observe, that, in the first case, since 

d — o, 
we must have a = oo in order that we may have 

a = t; 
and if we put a = t in the value of a, we obtain 

a = oo : 

also in the second case, since d = u and a = «, we must have 

a = i ; 
and if we put a = u in the value of a, we obtain 

a — I. 

Hence in using (i) we are using a formula for a which 
satisfies the two cases of carrying-strength and of primitive 
safe strength ; and the question as to its being a suitable value 
to use must depend upon how well it will satisfy intermediate 
values, i.e., cases where the two extremes are not those of the 
carrying-strength nor of the primitive safe strength. 



LAUNHARDT'S FORMULA. 43 1 

In the few cases of intermediate values experimented upon 
by Wohler, there is a very close agreement between the experi- 
mental results and those obtained by the formula. 

Now, put d = a — c in (1), and we have 

* - u , 
a = j-- a (a-c) (2) 

at — a 2 = at — au — c(t — u) 
.*. a 2 = au -f- (/ — #)<: 

•'• * = * + (/- *)-; (3) 

and if we denote by max L the greatest load on the entire 
piece, and by min L the least, we shall have 

c min L 



a max L 

Hence 

, , u N min Z . N 

a = u + {t - u) -, (4) 

max L 

this being in such a form as can be used. Or we may write it 
thus : 



!t — u min Z) 
1 + -r^ns> ( 5> 



this being the more common form. 

The values of the constants as determined by Wohler's 
experiments, and the resulting form of the formula for Phoenix 
axle-iron and for Krupp cast-steel, have already been given in 
§ 172. 

In the same paragraph are given the corresponding values 
of 5, the safe working-strength, the factor of safety being three. 



432 APPLIED MECHANICS. 

WEYRAUCH'S FORMULA FOR ALTERNATE TENSION ANA 
COMPRESSION. 

The constants used in this formula are : — 

i°. u, the primitive safe strength, which has been already 
defined. 

2°. s, the vibration safe strength ; i.e., the greatest stress 
per unit of area of which the piece can bear, without breaking, 
an unlimited number of applications, the other extreme stress 
which it bears alternately being —s. 

He lets a = greatest stress per unit of area, c = greatest 
stress of the opposite kind per unit of area. If a is tension, 
c is compression, and vice versa. 

Then, if d is the difference, 

d = a + c. 

Weyrauch writes, as before, 

a = ad, 

and gives to a a value which will satisfy the two special cases 
experimented upon ; viz., — 

i°. When a = u t c — o, d = u. 

2°. When a = s, c = s, d = 2s. 

He writes 

u — s 



2u — s — a 



a = 



2u — s — a 



d. (6) 



This value of a satisfies the two special cases referred to ; for 
in the first case, since d — n and a = u, we must have a = I ; 
and if we write a ■=. u in the value of a, we obtain a = I. Also 
in the second case, since d = 2s when a = s, we must have 
a = J ; and if we write a = s in the value of a, we obtain a = \. 

This, however, has not been tested for intermediate values. 



WEYRAUCH'S FORMULA. 43; 

Now write d = a + <; in equation (i), and we shall have 

u — s 
a = (a + r), 

2au — as — a 2 = au — as -f- (# — •*)*■ 

.*. a 2 = a» — (u — s)c (7) 



and if we write 



c 
\ a = u — (u — s)- ; (8) 



c max Z' 



# max Z* 



where max Z = greatest load on the piece, and max U = 
greatest load of opposite kind, so that, if Z is tension, L' shall 
be compression, and vice versa, we shall have 



, N max U 

--•-<"- ') wz' (9) 



this being in a form suitable to use, the more common form 
beinof 



!, u — s max Zf) , N 

1 4- -(.. (10). 

^ max Z ) 



The values of the constants as determined from Wohler's 
experiments, and the resulting form of the formulae for Phoenix 
axle-iron and for Krupp cast-steel, are given ( in § 176. 

The above demonstrations of Launhardt's and of Wey- 
rauch's formulae are substantially those given in the translation 
of Weyrauch's " Structures of Iron and Steel," by Professor 
Dubois, the explanations being somewhat changed. 



434 APPLIED MECHANICS. 



GENERAL REMARKS. 

In each case the value of a given by the formula (5) or (10) 
is the breaking-strength per unit of area for either tension or 
compression in Launhardt's formula ; and in the case of Wey- 
rauch's, the value of a is the breaking-strength per square inch 
of the same kind as max L ; i.e., tension if this is tension, or 
compression if this is compression. 

If either of these values of a be divided by 3, we have, 
according to Weyrauch, the safe working-strength. 

If, now, the piece be a tension piece, its area will be found 
by dividing its greatest load by the safe working-strength ; if, 
on the other hand, it be subjected to compression, and it be a 
short piece, its area will also be found by dividing the greatest 
compression by the safe working-strength per unit of area : but 
if it be a long column, and we wish to use Wohler's results, we 

must merely use the value - as the safe working-strength per 

3 
unit of area, and use this in whatever formula we may employ 

for calculating a column. 

WOHLER'S EXPERIMENTAL RESULTS. 

Wohler himself made his tests upon the extremes of fibre 
stresses of which a piece could bear, without breaking, an 
unlimited number of applications. He gives, as a summary of 
these results, the following : — 

In iron, — 

Between +16000 lbs. per sq. in. and —16000 .lbs. per sq. in. 
+ 30000 " " " o 

" +44000 " " " +24000 " " 

In axle-steel, — 

Between +28000 lbs. per sq. in. and —28000 lbs. per sq. in. 
" +48000 " " " o " " 

" +80000 " " " +35000 " " 



WOHLER'S EXPERIMENTAL RESULTS. 435 

In untempered spring steel, — 

Between +50000 lbs. per sq. in. and o lbs. per sq. in. 

" +70000 " " " +25000 " " 

" +80000 " « " +40000 " 

" +90000 *\ " " +60000 * " 

For shearing in axle-steel, — 

Between +22000 lbs. per sq. in. and —22000 lbs. per sq. in. 
« +38000 " " " o " " 

This table would justify the use, in Launhardt's and Wey- 
rauch's formulae, of the following values of u and s ; viz., — 
In iron, — 

u — 30000 lbs. per sq. in., 
s = 16000 lbs. per sq. in. 



In axle steel, 



u = 48000 lbs. per sq. in., 
s = 28000 lbs. per sq. in. 



In untempered spring steel, — 

u = 50000 lbs. per sq. in. 

And it would require, that if, with these values of u, and the 
values of / given in §§172 and 176, we put 

c = 24000 

in Launhardt's formula for iron, we ought to obtain approxi- 
mately 

a = 44000 ; 

and if we put c = 35000 in that for steel, we should obtain 
approximately 

a = 80000. 



43^ APPLIED MECHANICS. 



FACTOR OF SAFETY. 

We have seen that Weyrauch recommends, to use with 
Wohler's results, a factor of safety of three for ordinary bridge 
work and similar constructions. 

Wohler himself, however, in his " Festigkeits versuche mit 
Eisen und Stahl," says, — 

i°. That we must guard against any danger of putting on 
the piece a load greater than it is calculated to resist, by assum- 
ing as its greatest stress the actually greatest load that can 
ever come upon the piece ; and 

2°. This being done, that the only thing to be provided for 
is the lack of homogeneity in the material. 

3°. That any material which requires a factor of safety 
greater than two is unfit for use. This advice would hardly be 
accepted by engineers, however. 

He also claims that the reason why it is safe to load car- 
springs so much above their limit of elasticity, and so near 
their breaking-load, is, that the variation of stress to which they 
are subjected is very inconsiderable compared with the greatest 
stress to which they are subjected. 

GENERAL REMARKS. 

It is to be observed, — 

i°. The tests were all made on a good quality of iron and 
of steel, consequently on materials that have a good degree of 
homogeneity. 

2°. The specimens were all small, and the "repetitions of load 
succeeded each other very rapidly, no time being given for the 
material to rest between them. 

3°. No observations were made on the behavior of the piece 
during the experiment before fracture. 

4°. No experiments were made upon cast-iron and timber ; 
and the results of such experiments, if made, could hardly be 



SHEARING-STRENGTH OF IRON AND STEEL. 437 

expected to be of much value, as these materials are so lacking 

in homogeneity.. 

5 . As long as we are dealing only with tension, we can say 

without error that 

c _ min L , 

a max L ' 

but as soon as both stresses or either become compression, if 

the piece is long compared with its diameter, we cannot assert 

with accuracy that 

c max I! 



a max L * 

and hence results based on such an assumption must be to a 
certain extent erroneous. 

6°. When a piece is subjected to alternate tension and com- 
pression, it must be calculated so as to bear either : thus, if 
sufficient area is given it to enable it to bear the tension, it may 
not be able to bear the compression unless the metal is so dis- 
tributed as to enable it to withstand the bending that results 
from its action as a column. 

EXAMPLES. 

1. Determine the cross-section necessary for a wrought-iron tie 
where greatest load = 800000 lbs. and least load = 80000 lbs. 

2. Determine the cross-section necessary for a wrought-iron strut to 
bear the same loads. 

3. What is the greatest and what the least working- strength recom- 
mended by Wohler for wrought-iron? What for steel? Compare with 
ordinary methods, using factor of safety of four. 

§ 229. Shearing-Strength of Iron and Steel. — Some of 
the most common cases where the shearing resistance of iron 
and steel is brought into play are : — 

i°. In the case of a torsional stress, as in shafting. 



438 APPLIED MECHANICS. 

2°. In the case of pins, as in bridge-pins, crank-pins, etc. 

3°. In the case of riveted joints. 

In most cases the shearing is accompanied by tensions or 
compressions, or other stresses, and it is difficult to separate 
the effects ; so that, in the present state of our knowledge, we 
cannot explain the fact, that, when the breaking shearing- 
strength of wrought-iron or steel is deduced from an experiment 
on torsional strength, it is found to be about equal to the ten- 
sile strength, while when deduced from experiments on riveted 
joints, it is found to be about f or -| the tensile strength. 

Experiments on this subject are to be found in — 

i°. Steam-Boilers, by William H. Shock, U.S.N. 
2°. Kirkaldy : Experimental Inquiry into the Properties of Fagersta 
Steel. 

3°. A. B. W. Kennedy: Engineering. May 6, 1881. 

In regard to cast-iron, Bindon Stoney found the shearing 
and tensile strength about equal. 

As to shearing modulus of elasticity, Bauschinger's experi- 
ments on cast-iron give about two-fifths the tensile modulus of 
elasticity ; and Wohler's experiments gave for steel almost 
exactly two-fifths the tensile modulus, his value being 1 1236500 
when the tensile modulus was 29600000. According to the 
theory of elasticity, the modulus of elasticity for shearing 
should be about two-fifths that for tension or compression, as 
will be shown later ; and these experiments furnish a most 
beautiful confirmation of the theory of elasticity. 

The cases where shearing comes in play in" wrought-iron and 
steel will, therefore, be treated separately. 

§ 230. Torsional Strength of Wrought-iron and Steel. — 
The most common custom for computing the strength of a 
shaft has been to use one based upon its twisting-moment ; and 
hence using the shearing breaking-strength of the material, as 
determined from an experiment on simple torsion, for our con- 



TORSIONAL STRENGTH OF WROUGHT-IRON, ETC. 439 

stant. It is generally the fact, however, that, when shafting is 
running, the pulls of the belts create a bending backwards and 
forwards, bringing the same fibre alternately into tension and 
compression ; and this is combined with the shearing-stresses 
developed due to the twisting-moment alone. No tests have 
been made upon the effect of the combination of these stresses 
under working conditions ; and until we have a systematic set 
of tests of this character, we shall not be able to predict with 
certainty the behavior of a shaft under working conditions. 
At the two extremes of these general cases are : — 

i°. The case when the portion of a shaft between two 
hangers has no pulleys upon it, and when the pulls on the 
neighboring spans are not so great as to deflect this span appre- 
ciably. That is a case of pure torsion : and if the shaft is run- 
ning smoothly, with no jars or shocks, and no liability to have 
a greater load thrown upon it temporarily, we should properly 
compute it by the usual torsion formula, given in § 212 ; and we 
may use for breaking-strength of wrought-iron and steel the 
tensile strength, and use a factor of safety no greater than six. 

2°. The case when, pulleys being placed otherwise than near 
the hangers, the belt-pulls are so great that the torsion becomes 
insignificant compared with the bending, and then it would be 
proper to compute our shaft so as not to deflect more than 12V0" 
of its span under the load, or better, not more than ^Vo : °^ 
course we should compute also the breaking transverse load, 
and see that we have a good margin of safety. 

In other cases, experiments upon shafting under working 
conditions are needed; as up to the present time all such mat- 
ters have been decided in one of three ways, as follows : — 

i°. By using the ordinary torsion formula combined with a 
large factor of safety. 

2°. As recommended by D. K. Clark, by computing the 
shaft also for deflection, and providing that its deflection shall 
not exceed -J2V0 or TeVo °^ * ts s P an - 



440 APPLIED MECHANICS. 

This, however, neglects the torsion, and also the rapid 
change of stress upon each fibre from tension to compression, 
and vice versa. 

3°. By using the formula of Grashof or of Rankine for com- 
bined bending and shearing ; but this has not had the constants 
worked out from a test on such combined stress, and it is 
seldom used. 

In regard to experiments upon torsional strength, they are 
usually made by twisting a short piece of shaft until rupture 
occurs, measuring its twist under smaller loads, and from these 
computing the modulus of shearing elasticity. 

Such tests have been made by — 

i°. Kirkaldy : D. K. Clark's Rules and Tables. 
2°. D. K. Clark : English Civil Engineers' Committee. 
3°. Major Wade. 

4°. United-States Board to test Iron and Steel : Executive Docu- 
ment No. 23, 46th Congress, 2d session. 

5 . Professor Thurston : Materials of Engineering. 

The general result has usually been, to obtain about the 
same shearing breaking fibre stress as the tensile strength of 
the material, and a modulus of elasticity about two-fifths the 
tensile modulus. Of course the values vary more or less, some- 
times being greater and sometimes less than the values given 
above, according to the quality of the iron ; so that we can de- 
termine by experiment only what any one iron will bear. None 
of these tests will be recorded here, but some of the rules in 
common use for figuring the strength will be given : they are 
merely some of the formulae already referred to, with the con- 
stants changed. 

Unwin gives the direction, — 

i°. That the axle must be calculated as a beam to bear 
the weight of the pulley and the belt-pull that is to come 
upon it. 



TORSIONAL STRENGTH OF WROUGHT-IRON, ETC. 44 1 

2°. For shafting transmitting power, and subject to torsion 
only, he gives 

d = af^R = £y — , 

where d = diameter in inches, PR = twisting-moment in inch- 
pounds, N = number of turns per minute, HP = number of 
horses-power to be transmitted ; and he gives for a and j3 the 
following values : — 



Wrought- iron 
Cast-iron . 
Steel . . . 



0.08275 


3- 2 94 


0.1042 


4.150 


0.0723 


2.877 



|3. 



these values assumingior safe greatest fibre stress, for wrought- 
iron, 9000; for cast-iron, 4500; and for steel, 13500 lbs. per 
square inch. 

3 . For the crank-shaft of a steam-engine he advises us to 
consider the maximum twisting-moment as 1.3 times the mean. 

4 . For combined twisting and bending, he gives the Ran- 
kine formula, equation (2), § 215, and puts it in the following 
form : 



d= yz: Hx> + 1 tf 



*<ffT s 



where d = diameter in inches,/ = outside fibre breaking shear- 
ing-strength per square inch, T == twisting-moment in inch- 

= ratio of bending to twisting moment. 

Then, calling 



pounds, K = — 



n = 17 K + \/X 2 + i, 



44 2 APPLIED MECHANICS. 

n will be the ratio of the diameter to be used, to the diameter 
needed to resist the torsion alone ; and he gives, when 

K = 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 3.00, 
n = 1.09 1. 17 1.26 1.36 1.42 1.49 1.56 1.62 1.83. 

For the propeller shaft of a steam-vessel, he advises to use 

K — 0.25 to 0.50 .*. n = 1.09 to 1. 17. 

For line shafting in mills, he advises 

K — 0.75 to 1.00 /. n = 1.26 to 1.34. 

For crank-shafts and heavy shafting subject to shocks, 

K — 1. 00 to 1.50 .*. n = 1.34 to 1.49. 

5°. For forged crank-shafts, he gives 



^=4-55V N 

To control the deflection of shafting 10 or 11 feet between 
hangers, he gives 

v, 



@ 



where L = span in inches, and y = 54 to 60. 
Professor Thurston gives, — 
For head shafts well supported against springing, — 



Wrought-iron, d = ^ I2 $ HF . Cold-rolled iron, d = \ 1S ^ 
For line-shafting, hangers 8 feet apart, — 



Wrought-iron, d = ^W*£ Cold-rolled iron, d = ^ *$**?. 

For transmission simply, no pulleys, — 



Wrought-iron, d = y/ 62 -5^ Cold-rolled iron, d = \ ~^~- 



BRIDGE-PINS. 



443 



Mr. James B. Francis gives the following table for distance 
between bearings of shafting carrying no side strain : — 



Diameter. 


Wrought-Iron. 


Steel. 


Diameter. 


Wrought-Iron. 


Steel. 


in. 


ft. 


ft. 


in. 


ft. 


ft. 


2 


i5-5 


15-9 


6 


22.3 


22.9 


3 


17.7 


18.2 


7 


23-5 


24.I 


4 


19.5 


20.0 


8 


24.6 


25.2 


5 


20.9 


21.6 


9 


25-5 


26.2 



It should be observed, that it is only a mild steel that can 
be used for shafting, — steel of about 60000 lbs. tensile strength 
per square inch : hence the calculations cannot differ very 
greatly from those for wrought-iron. 

§ 231. Bridge-Pins. — In the case of a bridge-pin or other 
pin, the shearing-stress is always accompanied by a very large 
element of bending ; so that the principles of transverse 
strength come into play very largely, and perhaps in many 
cases wholly. 

The rules for proportioning bridge-pins used by different en- 
gineers and constructors have been various, but have generally 
been founded upon the theory of transverse strength ; for those 
given by Charles Bender, the student is referred to his treatise 
on "Bridge-Pins" (Van Nostrand's Science Series). No attempt 
will be made to give any of the rules used ; but, inasmuch as 
there were quite a number of such pins tested at the Watertown 
Arsenal, there will be given a summary of the tests, which are 
to be found in detail in Executive Document 12, 47th Congress, 
first session. All the pins tested were of wrought-iron, and 
had semicircular seats at middle and ends. 

From this table the modulus of elasticity can be figured, 
and the deflection of the pin under any given load and span ; 



444 



APPLIED MECHANICS. 



or the table may be employed to interpolate, to find the sizes 
to be used without causing: too much deflection. 



Diameter, 
in inches. 


Clear 

Span. 


Width 

of 
Middle 
Bear- 
ing. 


Load nearest 

one-half 

Maximum 

Load, in lbs. 


Deflection, 
in inches. 


Maximum 
Load ap- 
plied, in lbs. 


Deflection, 
in inches. 


Amount of 

Indentation, 

in inches. 


2.500 


24 


I 


I OOOO 


O.0517 


2000O 


I.270O 


O.OO70 


2.497 


24 


I 


8000 


O.0420 


15000 


O.329O 


0.0020 


2.500 


18 


I 


I OOOO 


O.O238 


2IOOO 


O.1660 


O.OO25 


2.5OO 


18 


I 


I OOOO 


O.0248 


18000 


O.IOOO 


< 0.00 10 


2.500 


12 


- 


I50OO 


O.OI33 


30000 


O.O367 





2.500 


12 


- 


I4OOO 


O.OI29 


2700O 


O.O437 


< 0.00 10 


2.500 


6 


I 


33OOO 


O.O066 


6500O 


O.O263 


0.0030 


2.500 


6 


I 


30000 


O.O064 


60000 


O.O184 


- 


3.OOO 


24 


- 


1 1 OOO 


O.0278 


22000 


0.1260 


- 


3.OOO 


24 


- 


1 1000 


O.0278 


22000 


O.I23I 


- 


3.OOO 


12 


I ft 


25000 


O.OIOI 


50000 


O.O414 


- 


3.000 


12 


if 


25000 


0.0102 


50000 


O.O464 


- 


3.000 


6 


- 


62000 


0.0077 


125000 


O.O24O 


- 


3.OOO 


6 


If 


60000 


0.0070 


I25000 


O.O25O 


- 


3495 


24 


If 


20000 


0.0302 


40000 


O.I68I 


- 


3495 


24 


If 


20000 


0.0310 


35000 


O.O923 


- 


3495 


12 


- 


40000 


O.OI2I 


80000 


O.O33O 


- . 


3495 


12 


- 


40000 


O.OI24 


800OO 


O.O34O 


- 


3496 


6 


- 


90000 


O.O083 


1700OO 


0.0222 


0.0040 


3496 


6 


- 


90000 


O.O087 


160OOO 


O.OI7I 


- 


4.000 


24 


If 


25000 


0.0227 


5OOOO 


O.0667 


- 


4.000 


24 


If 


25000 


O.O234 


50000 


O.0682 


- 


4.000 


J 2 


If 


55000 


O.OO97 


I I OOOO 


O.O258 


- 


4.000 


12 


If 


55000 


O.OI20 


I I OOOO 


O.O255 


- 


4.500 


24 


If 


38000 


O.O24O 


75000 


O.0648 


- 


4.500 


24 


If 


40000 


O.O260 


75000 


O.063O 


- 


4.500 


12 


- 


80000 


O.OO99 


160000 


O.O245 


- 


4.500 


12 


- 


80000 


O.OO96 


170000 


O.O281 


- 


4-877 


24 


If 


50000 


O.O24O 


I 00000 


O.O912 


- 


4.877 


24 


If 


40000 


O.OI9O 


85000 


O.O460 


- 


4.877 


12 


If 


I 00000 


COI55 


200000 ' 


C.O324 


- 


4.877 


12 


x 4 


I 00000 


O.OI52 


200000 


O.O322 


— 



RIVETED JOINTS. 445 



§ 232. Riveted Joints. — The most common way of uniting 
plates of wrought-iron or steel is by means of rivets. It is, 
therefore, a matter of importance to know the strength of such 
joints, and also the proportions which will render their efficien- 
cies greatest ; i.e., that will bring their strength as near as 
possible to the strength of the solid plate. 

In § 177 was explained the mode of proportioning riveted 
joints usually taught, based upon the principle of making all 
the resistances to giving way equal, and assuming, as the modes 
of giving way, those there enumerated. This theory does not, 
however, represent the facts of the case, as — 

i°. The stresses which resist the giving- way are of a more 
complex nature than those there assumed, so that the efficiency 
of a joint constructed in the way described above may not be 
as great as that of one differently constructed ; 

2°. The effects of punching, drilling, and riveting, come in 
to modify further the action ; and 

3 . The purposes for which the joint is to be used, often fix 
some of the dimensions within narrow limits beforehand. 

In order to know, therefore, the efficiency of any one kind 
of joint, we must have recourse to experiment. And here again 
we must not expect to draw correct conclusions from experi- 
ments made upon narrow strips of plate riveted together with 
one or two rivets ; but we need experiments upon joints in wide 
plates containing a sufficiently long line of rivets to bring into 
play all the forces that we have in the actual joint. The greater 
part of the experiments thus far made have been made upon 
narrow strips, with but few rivets, — there have been but few 
of the latter class of tests, — and no complete and systematic 
series of tests has thus far been carried out, though such a 
series has been begun on the government testing-machine at 
the Watertown Arsenal. 

The only tests that will be quoted here, are : — 

i°. The summary of the tests of this series that have been 



446 APPLIED MECHANICS. 

made thus far, and are recorded in Executive Document No. I, 
47th Congress, second session, and Executive Document No. 5, 
48th Congress, first session. 

2°. A few tests made recently by David Kirkaldy. 

While it is from these tests upon long joints that we can 
derive correct and reliable information to use in practice, and 
hence while the experiments already made give us some infor- 
mation in regard to such joints as were tested, nevertheless, as 
these tests have not yet been carried far enough to furnish 
all the information we need, and to settle cases that we are 
liable to be called upon to decide, therefore, before quoting 
the above experiments, some of the rules and proportions in 
use in practice at the present time, and the modes of deter- 
mining them, will be first explained. 

In this regard we must observe that practical considerations 
render it necessary to make the proportions different when the 
joint is in the shell of a steam-boiler, from the case when it is 
in a girder or other part of a structure. 

In the case of boiler-work, the joint must be steam-tight, and 
hence the pitch of the rivets must be small enough to render 
it so : whereas in girder-work this requirement does not exist ; 
and hence the pitch can, as far as this requirement goes, be 
made greater. 

It is probable, that, with good workmanship, we might be able 
to secure a steam-tight joint with considerably greater pitches 
than those commonly used in boiler-work ; and now and then 
some boiler-maker is bold enough to attempt it. 

Tables of usual dimensions employed and recommended by 
Robert Wilson, Thomas Box, and Unwin respectively, will now 
be given. 

The following tables give the proportions recommended by 
Robert Wilson for boiler-work, and by Thomas Box for girder- 
work : — 



RIVETED JOINTS. 



447 



PROPORTIONS GIVEN BY ROBERT WILSON FOR USE IN BOILER-WORK. 





Diameter of 




Pitch of Rivets. 


Double-Riveted Butt-joints 


Thickness 
of Plates. 


Rivets. All 
Lap-joints and 


Pitch of Rivets. 
Single-Riveted 


Double-Riveted 
Lap-joints, and 


with Double Strips. 






Butt-joints with 


Lap-joint. 


Butt-joints with 


Diameter of 


Pitch of Rivets. 




Single Strips. 




Single Strips. 


Rivet. 




i 


i 


li- 


It 


_ 


_ 


A 


f 


lt 


2i 


- 


- 


t 


H 


If 


2i 


t 


2? 


A 


2. 

4 


I* 


2i 


t 


2f 


t 


1 


I* 


4 


f 


2f 


A 


i 


2i 


2f 


H 


2^ 


t 


I 


2i 


2| 


! 


3 


H 


I 


2i 


2| 


f 


3i 


5. 

4 




at 


3i 


1 


3i 


if 




2i 


3i 


8 


3i 


1 




2i 


3i 


I 


3» 


if 


T L 


2i 


3* 


I 


-3 a 
J4 


i 


T l 


2Y 


3i 


I* 


4 



He gives, for the lap in single riveting, 3 times the diameter 
of the rivet, and never more than 3.3 times. 



PROPORTIONS RECOMMENDED BY THOMAS BOX FOR USE IN GIRDER- 
WORK. 



Single-Riveted Lap-J 


OINTS. 




Double-Riveted Joints. 


Thickness 


Diameter 


Pitch of 


Lap. 




Thickness 


Diameter 


Pitch of 


Lap. 


of Plates. 


of Rivets. 


Rivets. 




of Plates. 


of Rivets. 


Rivets. 


1. 

8 


A 


I* 


I 


3 

8 


tt 


2ft 


2\ 


A 


A 


I* 






A 


1 


3 


3 


1. 
4 


i 


I* 


If 




i 


it 


3A 


3i 


A 


A 


If 


I* 




A 


J 

8 


3i 


31 


3. 

8 


tt 


2 


If 




f 


if 


3! 


3f 


A 


2. 

4 


2* 


i* 




tt 


itV 


3if 


4* 


i 


if 


21% 


2 




1 


li 


4A 


4i 



448 APPLIED MECHANICS. 

i°. In regard to the diameter of the rivets, Robert Wilson 
first shows, that by taking the crushing-strength in front of the 
rivet at 89600, and the shearing-strength of the rivet at 47040 
lbs. per square inch, we should find that the joint would be safe 
against crushing the metal in front of the rivet with a diameter 
of rivet equal to 2 \ times the thickness of the plate ; he recom- 
mends the diameters given in the table, which for plates \ inch 
and T 5 g inch thick are double the thickness of the plate (this 
being a rule frequently used), and for thicker plates they grow 
gradually less. 

Thomas Box, in deducing his diameters for girder-work, gives 
the formula 

d = \t + A, 

from which he calculates the diameters given in his table, giv- 
ing this as an empirical rule. 

2 . In regard to the pitch of the rivets, Robert Wilson, by 
calling the shearing-strength of the rivets per square inch equal 
to the tensile strength of the plate per square inch, deduces the 
formula 

/ = — - + a. 

4/ 

The values recommended by him differ, however, somewhat 
from the results of this formula, in order to retain a larger 
section of plate between rivets. 

Thomas Box deduces his values of the pitch by considering 
the shearing-strength of the rivet per square inch as equal to 
I the tensile strength of the plate per square inch, and then 
calculating the joint so as to give equal strength for tearing 
and 'shearing. 

3 . In regard to the lap, Robert Wilson computes it so that 
there shall be strength enough to resist breaking through : his 
formula has been given in § 177. This would give, for the lap, 
the formula for single riveting, 

/ = o.Zid. 



RIVETED JOINTS. 449 



Hence he concludes that the common rule of making the dis- 
tance between the hole and edge of the plate equal to the 
diameter of the rivet is to provide sufficient resistance in this 
regard : this rule, however, he contradicts by his empirical rule 
given just after the table. 

Thomas Box, on the other hand, gives a graphical construc- 
tion for the lap, which practically accounts the shearing and the 
tensile strength of the plate per square inch the same, and pro- 
vides sufficient strength to prevent the rivet from shearing out 
the plate in front of it. His results are a little larger than 
three times the diameter of the rivet. 

Next, as to Unwin's recommendations, — 

i°. In regard to the diameter of rivet, he says that the 
diameters used in practice range from 

d = \t + I 

to 

d = \t + f, 

and recommends, as a good rule to follow, 

d= i.2\[t. 

It will be seen that he is thus recommending a little larger 
diameters than Wilson or Box. 

2 . As to the pitch, he determines it from the formula 

{p-d)tf t = ^d 2 f, 
4 

where/ = pitch, / = thickness of plate, d = diameter of rivet, 
f t = tensile strength of plate between rivet-holes after the rivet- 
ing has been done,/^ = shearing-strength of rivet per square 
inch, using such values of f and f s as he considers suitable. 
The result of all this will be shown in the following tables, 
given by Unwin. 



450 



APPLIED MECHANICS. 



Ratio of Tearing and Shearing Stre?igthj in Riveted Joints, f t being Tensile 
Strength per Square Inch after the Plate has been Punched or Drilled. 





Iron Plates, Iron Rivets. 


Steel Plates, Steel Rivets. 


Drilled or 
Punched, 
and An- 
nealed or 
Rymered. 


Punched. 


Drilled or 
Punched, 
and An- 
nealed or 
Rymered. 


Punched. 


Single-riveted 

Double-riveted 

Treble-riveted 


O.94 
I.02 
I.05 


O.77 
O.85 


1.26 
1-34 
1.36 


I.05 
1. 17 



Values of Pitch for Single Riveting for Various Values ofj- 



3. 



] b 
2. 

I 



Ik 



O.72 
O.78 
O.85 
O.92 
O.98 
1. 10 

I..I7 

I.30 



Single Rivetim 



Iron Rivets. 



Iron Punched 
Plates. 



Iron Drilled or 

Punched, and 

Annealed or 

Rymered 

Plates. 



Steel Rivets. 



Steel Punched 
Plates. 



Steel Punched 

and Annealed, 

or Rymered 

Plates. 



Pitch for values of <- 
fs 



O.75 O.85 O.95 



I.05 1. 15 I.25 



•35 



2.46 


2.25 


2.09 


2.02 


1.96 ' 


1.85 


i-77 


2.48 


2.2S 


2.12 


2.06 


1.99 


1. 89 


1.81 


2.58 


2.38 


2.22 


2.15 


2.09 


1.98 


1.90 


2.69 


2.48 


2.32 


2.25 


2.19 


2.08 


2.00 


2.69 


2.40 


2.25 


2.19 


2.13 


2.03 


i-95 


2.79 


2-59 


2-43 


2-37 


2.31 


2.20 


2.12 


2.81 


2.62 


2.46 


2.40 


2-34 


2.24 


2.16 


3-07 


2.86 


2.70 


2.63 


2.56 


2-45 


2.36 



1.69 

1.72 
1.81 

1.90 

1.87 

2.04 

2.08 
2.28 



RIVETED JOINTS. 



451 



By "real diameter" he means the diameter after riveting. 









Double Riveting. 


Iron Rivets. 




Steel Rivets. 


V 

rt 


> 

5 


! 
s 














Iron 






Steel 


- 04 


hi 


*s 


Iron 


• Drilled or 




Steel 


Drilled or 





« 


i) 


Punched 


Punched, 




Punched 


Punched, 


u 


E 

rt 


E 


Plates. 


and Ry- 




Plates. 


and Ry- 


'J3 
H 


Q 

"rt 

a 

'g 



rt 

Q 
73 




mered Plates. 






mered Plates. 




Pitch of rivets for value ^ — 










O.85 


I.OO 1. 10 1.20 


i-35 


g 

TF 


tt 


O.72 


373 


3-33 


3.12 


2.91 


2.66 


3. 

8 


4 


O.78 


378 


3-33 


3.12 


2.91 


2.66 


A 


H 


O-85 


3-91 


345 


3- 2 4 


3-03 


2.77 


£ 


1 


0.92 


4-05 


3.58 


3-37 


3.16 


2.88 


5. 

8 


If 


O.98 


3.82 


3-39 


3.18 


3.OO 


2.76 


5. 

4 


nV 


1. 10 


4.08 


3-^3 


3-42 


3.22 


2.98 


I 
8 


if 


1.17 


4.06 


3-t>3 


342 


3-23 


2.99 


1 


ii 


1.30 


4.42 


3-95 


374 


3-52 


3.26 i 



It will be noticed, that, having used a larger rivet than Wil- 
son or Box, he naturally used a larger pitch. 

For lap, he gives the following values, computed by the same 
rule as Wilson computes his, but with a different constant ; and 
he then compares them with values of 1.5^, which he states to 
be an ordinary rule. 



d 


i 


£ 


I 


Z. 

8 


1 


i| 


J 4 


For iron, / = 


1.00 


1. 14 


I.29 


I.4I 


i-55 


1.67 


I.80 


For steel, / = 


0.86 


O.98 


1. 12 


1.22 


i-35 


1.46 


l -S7 ; 


l.$d 


o75 


O.94 


1. 12 


i-3i 


1.50 


1.69 


1.88 



452 



APPLIED MECHANICS. 



Efficiency of the yoint. — The efficiency of the joint is an 
item of great importance. Sir William Fairbairn's experiments 
gave the following efficiencies : — 

Entire plate ioo per cent, 

Double-riveted joint 76 " 

Single-riveted joint 56 " 

and these efficiencies have been very much quoted and used. 

Mr. Forney, in his " Catechism of the Locomotive," gives 
the following : 

Solid plate . , 100.0 

Single-riveted seam, punched holes, ^ rivets .... 50.5 

" drilled " £ " .... 60.0 

Double-riveted seam, drilled and zigzag holes, f rivets . 70.0 

" " " " " " 7. " 72 o • 

Single riveted seam, punched holes, with covering- plate . 65.3 

On the other hand, Unwin gives the following tables for the 
efficiencies in different cases : — 



(fl 


■a 

1 


!> 


Single Riveting. 




Iron Drilled or 




Steel Punched 


JS 


*o 


« 


Iron Punched 


Punched, and 


Steel Punched 


and Annealed, 


<— 


4> 





Plates. 


Annealed or 


Plates. 


or Rymered 




E 


« 




Rymered Plates. 




Plates. 


c 


rt 

s 


s 
s 


















B 


C 

*£ 



Q 
u 




Efficiency of joint 


s for values of <1 
fs 


= 




£ 




O.77 


O.88 


O.9 


1.0 


-h 


tt 


O.72 


0-55 


0.52 


O.58 


0.56 


0.57 


o-S5 


o-59 


o-57 


f 


3. 

4 


O.78 


°-53 


0.51 


°-55 


0.54 


°-55 


o-53 


0.57 


o.55 


A 


if 


O.85 


0.52 


0.49 


o-55 


0.54 


o-53 


0.51 


o-55 


0-53 


\ 


\ 


O.92 


0.51 


0.49 


0.52 


0.52 


0.52 


0.50 


0.54 


0.52 


b. 

8 


H 


O.98 


0.48 


0.45 


0.49 


0.48 


0.49 


0.47 


0.50 


0.48 


f 


iA 


I. IO 


0.47 


0.44 


0.48 


0.47 


0.47 


045 


0.48 


0.46 


7 
8 


1* 


I.I7 


o-4S 


0.42 


0.46 


0.45 


0-45 


043 


0.46 


0.44 


I 


*i 


I.30 


0.42 


0.40 


0.46 


045 


0.45 


o-43 


0.45 


043 



RIVETED JOINTS. 



453 



V 


> 

3 


> 


Double Riveting. 


Iron 


Iron Drilled 
or Punched, 




Steel 


Steel Drilled 
or Punched, 


E 


o 

hi 




Punched 


and 




Punched 


and 


o 




<L) 


Plates. 


Ryniered 




Plates. 


Rymered 


c 

IS 

H 


■E 
5 

"rt 

c 

s 

o 
5? 


s 

5 
7a 
B 




Plates. 






Plates. 




Efficiency of joints for values of *— = 
fs 








O.85 


0.95 I. OO 1. 00 


I.06 


A 


H 


O.72 


O.69 


O.74 


0.77 


0.75 


O.77 


8 


1 


O.78 


0.68 


0.73 


075 


0-73 


0-75 


f<> 


13 


O.85 


0.66 


O.7I 


0.74 


O.72 


O.74 


t 


i 


O.92 


0.65 


O.7O 


0.73 


O.71 


0-73 


t 


W 


O.98 


0.63 


O.67 


0.69 


O.67 


O.69 


1 


i* 


I. IO 


0.62 


0-66 


0-68 


O.66 


O.68 


I 
« 


yi 


I.17 


0.60 


0.65 


0.66 


O.64 


O.65 


1 


ii 


I.30 


0-60 


0-63 . 


0.65 


O.63 


O.64 



Punching and Drilling. — One matter in this connection 
that has occupied a good deal of attention is the relative ad- 
vantages of punched and drilled holes. The usual practice is 
to punch the holes, and it is less expensive than drilling them. 

On the other hand, it is generally acknowledged, and has 
been shown by a number of experimenters, that punching in 
most cases injures the metal around the hole: the amount of 
this injury may vary from a very small quantity up to 20 per 
cent in iron, and up to 35 per cent in steel, plates. The amount 
of injury will vary according to the quality of the plate, and 
also according to the amount of clearance between the punch 
and the die ; the injury being less in plates of good quality and 
ductile, and greater in hard and brittle plates, also less when 
the clearance between the punch and the die is ample than 
when it is too small. 



454 APPLIED MECHANICS. 

Accounts of tests on this subject are to be found in — 

Sir William Fairbairn : Application of Iron to Building Purposes. 

J. Barba : Use of Steel in Construction. 

A. F. Hill : Paper in the Proceedings of Society of Engineers of Western 

Pennsylvania. 
Executive Document No. i, 47th Congress, 2d session, p. 142 et seq. 
Executive Document No. 5, 48th Congress, 1st session. 

There are two other matters that sometimes appear to 
modify these statements as far as breaking-strength is con- 
cerned ; i.e., — 

i°. When the specimen is a grooved one, as it must be 
when punched or drilled, its apparent ultimate strength in the 
testing-machine is greater than it would be if it had any oppor- 
tunity to stretch. 

2°. Cold-punching has, to a small extent, a similar effect to 
cold-rolling ; i.e., it may harden the metal a little, and increase 
its breaking-strength on this account, while rendering it less 
ductile, and hence more brittle. 

The injury done by punching may be almost entirely re- 
moved in either of the following ways : — 

i°. By annealing the plate. 

2°. By reaming out the injured portion of the metal around 
the hole ; i.e., by punching the hole a little smaller than is de- 
sired, and then reaming it out to the required size. This 
removing the injurious effect of punching is more needed in 
steel than in iron plates. 

There is a certain friction developed by the contraction of 
the rivets in cooling, tending to resist the giving-way of the 
joint ; but this is likely to disappear before fracture takes place, 
and cannot, therefore, be depended upon. 

The shearing-strength of the rivets would appear to be 
about -| or -| the tensile strength of the plate. 



RIVETED JOINTS. 455 



Government Experiments. — These experiments form, as has 
been already stated, the first portion of a systematic series ; and 
the tables of results are given here, because it seems to the 
author, that, although the series is not yet completed, yet these 
tests themselves furnish more reliable information in regard to 
the behavior and the strength of joints made somewhat in these 
proportions than any other experiments that have been made, 
and that the figures themselves furnish the engineer with the 
means of using his judgment in many cases where he had no 
reliable data before. 

Thus it is plain, that, when we compute the crushing-strength 
in front of the rivet, it is not this direct crushing action that is 
produced, as the small piece of plate that lies in front of a rivet 
would not bear so much unless re-enforced by the metal on the 
sides. 

Nevertheless, the compression per square inch on the bear- 
ing-surface of the rivet (i.e., the pull on one rivet divided by 
the longitudinal projection of its bearing on one plate, however 
it be resisted) apparently forms an important element in the 
strength of the joint, and shows the advantage of large rivets ; 
this being one of the deductions made by Mr. James E. Howard 
(who conducted the tests) from these tests, inasmuch as some 
of the experiments tried with the same section of plate between 
the holes, and with rivets of different size, showed a greater 
efficiency for the larger rivets. 

In calculating the resistance to tearing out of the plate in 
front of the rivet by the rules of Wilson or Unwin, we must 
observe that there is not the action that is considered in the 
formulae. 

The forces brought into play are more complex, and it is 
only experiment that can show us what they are. 

A perusal of the tables will give a good idea of the shear- 
ing-strength per square inch of the rivet iron, which is seen to 
be less than the tensile strength of the solid plate ; also the 



456 APPLIED MECHANICS. 

loss of strength of the plates due to the entire process of 
riveting, punching, drilling, and driving the rivets ; also the 
efficiencies of the joints tested. 

One of the strongest single-riveted joints tested was a single- 
riveted lap-joint with a single covering-strip. 

The apparent anomaly of the punched plates in a few cases, 
showing a greater strength than the drilled plates, is explained 
by Mr. Howard to be due to the strengthening effect of cold- 
punching combined with smallness of pitch, inasmuch as then 
the masses of hardened metal on the two sides re-enforce each 
other. 

Further than this, the student is left to study the figures 
themselves as to the effect of different proportions, etc. 

In regard to these tests, it is stated in the report that — 

r°. "The wrought-iron plate was furnished by one maker 
out of one quality of stock." 

2°. " The steel plates were supplied from one heat, cast in 
ingots of the same size ; the thin plates differing from the 
thicker plates only in the amount of reduction given by the 
rolls." 

The modulus of elasticity of the metal was, iron plate, 
31970000 lbs. ; steel plate, 28570000 lbs. 

In the tabulated results, the manner of fracture is shown 
by sketches of the joints, and is further indicated by heavy 
figures in columns headed "Maximum Strains on Joints, in lbs., 
per Square Inch." 



RIVETED JOINTS. 



457 



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458 



APPLIED MECHANICS. 



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RIVETED JOINTS. 



459 









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RIVETED JOINTS. 477 



The following tests, made recently by David Kirkaldy, are 
also quoted from Executive Document No. 48, 47th Congress, 
second session, p. 24 et seq. 

These tests were made on joints of J-inch steel plate. The 
rivet-holes of 1418, "4," 1406, " 1," 1410, "2," and 1414, "3," 
were drilled separately with a 1 J^-inch drill. The plates were 
then heated, as the boiler-plates themselves would be for roll- 
ing, and, after cooling, were put together, and the holes reamed 
out to I -J inch. 

The rivet-holes of the remainder were punched separately 
with a J-inch punch on a iJg-inch die. The plates were then 
treated like the others. All samples were riveted together the 
full width of the plates, 24 inches. 

The efficiencies deduced from the Watertown experiments 
are not summed up here, as they can be obtained from the 
tables. 

On page 480 will be given the tests of grooved specimens 
of wrought-iron and steel made at the arsenal on plates used 
in the joints already referred to. These plates include both 
punched and drilled plates, and vary in thickness from \ inch 
to 1 inch. 



478 



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RIVETED JOINTS. 



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E 




r-^ oo r^ 












\0 ro 


<U 










Tf 


"-> fo 




u-1 




p 




"1- vO 


>>3 




VO 






















3 ^ 

£1 


c3 
3 
23 


O 
X 




t jj 


8 


vO 

o 

O O w 

LT> LT) (JO 








o 


VO ro 
00 lo 



vo Cfl 



VO 



3 

o £ 



o 

oo 
oo 
d 

4) 
| X VO 

— re . 

3 !_, ro m 



P K? 




O O >o rt b " « 

m lo j>, oj -fs ,-. 

o i-i +-> r3 <u y 

to c-> re w <u ^ 



o 




+j 


w 


to 


o o 




r^, 


rf 3 




r^ 


^ n 




r^ 




ro 


vO <u 



P4 



K 


oo 


N £ 


oo 


5 ,0 


o 


3 


<u 


£3 


S X 

'-5 a 



£~s 



o „ 

OO 7) 

^00 « 

oo js 



5 






^ 



s? 2 8 



T! -5 



S c ^- ^2 " 

»o o O > 4) 

3 ,o, a •• •• 

g S re rt rt 

h AS 5 5 



tf 












<u 




"S 






QJ 




"d 








re 




> 


. > 


01 














3 
o- 














o 

00 






,£3 

6 




O 


*o 


r5 


O 










3 
3 
T3 


O 


* 


X! 

3 


J3 
u 




M-l 

o 




ft 


rt 




3 
re 






(-i 




4j 


,3 







i^H 






re 
o 




re 

3 


. re 


c^ 






o 


12 




<u 


re 




cr 1 


3 
cr 1 


OJ 






3 


o 








• 




• 73 


n 


u 

3 




U 

i- 


7) 

o 






re 
3 
O 








3 


to 

a 


• 


3 

cr 

7} 

u 


c 

o 


t 

r 

a. 
it 


re 

•-5 




IT) 
O 


oX) 

3 


• in 

<n 

. 4) 

. ^ 


U 

7) 

J 

re 


CD 

— 


3 

jo 


OO 

U 
u 

W 


.2 
re 


{3 


> 

2 


43 


Sheari 

area 

Tensil 



480 



APPLIED MECHANICS. 



GOVERNMENT TESTS OF GROOVED SPECIMENS. 



Tensile Tests of A-in. 


Grocn 


'ed Specimens 


W 


rought-Iron 




Punched. 


E 




•S^ 












c a 


o . 




U 1 - 1 


« £ 





Jj 


-fi 


in 


w d- 






e ** 


P's 


r 


S8.9 


^ 


& 


Inch. 


Inch. 




0.48 


0.240 


48090 


0.46 


0.235 


46940 


0.46 


0.241 


49280 


0.49 


0.240 


55340 


0.44 


0.239 


51520 


0.47 


0.241 


49910 


0.97 


0.247 


49540 


0.98 


0.247 


49960 


0.94 


0.249 


50128 


0.96 


0.248 


46900 


0.98 


0.250 


46980 


0.96 


0.251 


46350 


1.47 


0.250 


37636 


1.50 


0.252 


37326 


1.48 


0.249 


41030 


1.48 


0.247 


39480 


1.47 


0.250 


37446 


i-45 


0.251 


39533 


1.96 


0.281 


43194 


1-95 


0.274 


47490 


1.95 


0.282 


41360 


1.92 


0.279 


43080 


2.03 


0.250 


41140 


1.99 


0.248 


39575 


2.42 


0.280 


36210 


2.40 


0.245 


42245 


2.47 


0.243 


42233 


2.46 


0.285 


42712 


2.48 


0245 


38125 


2.44 


0.248 


41620 


2.97 


0.247 


38964 


2.98 


0.241 


41540 


2.96 


0.241 


39972 


2.92 


0.240 


41712 


2.98 


0.250 


40430 


2-95 


0.247 


40850 



Tensile Tests of J-in. 


Grooved Specimens 


Wrought 


Iron 


Drilled. 


S 




•£-c 


















Z~ 


pa « 






(/> 


m a 1 






vin u,- 


■eO 


-3 * 


rt -° 

E ^-~ 


rs-s 


2 p. 


2 S.S 


^ 


H 


P 


Inch. 


Inch. 




0.51 


0.249 


55787 


0.52 


0.245 


55905 


0.52 


0.275 


57480 


0.52 


0.276 


56000 


0.49 


0.248 


49600 


0.50 


0.248 


56700 


0.47 


0.275 


54880 


0.51 


276 


57800 


1. 00 


0.276 


54300 


1.02 


0.273 


57700 


1. 00 


0.276 


53800 


1. 00 


0.280 


52430 


1. 00 


0.252 


49400 


1.02 


0.275 


54060 


1. 01 


0.247 


52770 


1. 00 


0278 


54600 


1.50 


0.276 


49130 


1.52 


0.273 


51300 


1.48 


0.251 


47220 


1. 51 


0273 


53400 


1.52 


0.275 


54180 


1.50 


0.276 


54600 


1.48 


0.274 


56250 


1.50 


0.249 


46260 


2.01 


0.275 


45900 


2.05 


0.279 


46820 


2.00 


0.275 


47950 


2.00 


0.278 


49640 


2.00 


0.286 


44650 


2.00 


0.275 


50780 


2.02 


0.279 


48850 


2.00 


0.277 


49840 


2-51 


0.244 


44980 


2.52 


0.280 


40150 


2.51 


0.282 


43150 


2.50 


0.244 


45500 


2.51 


0.285 


46500 


2.49 


0.242 


49520 


2.49 


0.242 


- 


2.50 


0.280 


44780 


3.02 


0.250 


45700 


3.02 


0.249 


44870 


3.00 


0.240 


46760 


3.00 


0.250 


45700 


2-93 


0.242 


47950 


2.99 


0.250 


48740 


2.98 


0.279 


45900 


3.01 


0.281 


44410 



Tensile Tests of a-in. 


Grooved Sp 
Steel PI 


ecimens 


ate 


Punched. 


E 




■£* 













c 3 


.: 




Jji—I 


« £ 





iz 





</> 


m ry 






H^ m 


^O 


C V 


rt ,0. 
E >-"" 


:s«s 


:a£ 


■= &C 


£ 


H 


P 


Inch. 


Inch. 




0.49 


0.250 


65120 


0.47 


0.249 


67010 


0.48 


0.249 


63420 


0.48 


0.248 


66550 


0.48 


0.247 


67060 


o.47 


248 


65300 


0.99 


0.249 


59840 


1. 00 


0.250 


62160 


1. 01 


0.249 


68246 


0.96 


0.250 


67330 


0.96 


0.248 


65966 


o.95 


0.245 


62700 


i-45 


0.248 


64080 


1.45 


0.252 


64000 


1.45 


0.249 


61025 


1. 51 


0.251 


59420 


1.96 


0.250 


59900 


i-93 


0.252 


63500 


1.98 


0.250 


59350 


1.96 


0.251 


59o6o 


2.49 


0.249 


581OO 


2.47 


0.249 


63900 


2-43 


0.250 


61640 


2-95 


0.251 


56530 


3.01 


0.249 


58780 


3-04 


0.253 


55500 


2.97 


0.252 


60060 


2.98 


0.251 


54050 


2.97 


0.249 


56040 



1 

Tensile Tests of 4-in'. 


Grooved Specimens 


Steel Plate 


Drilled. 


E 




■Sjs 












c c 


„; 






a £ 







*j 


t/> 


w 6> 


« S 




B^ <n 


*0 


c U 


rt jQ 

E ^~ 


P 


'rS 


a e-S 


H 


P 


Inch. 


Inch. 




0.52 


0.246 


67890 


0.54 


0.248 


67160 


0.53 


0.247 


66870 


0.50 


0.247 


65610 


0.51 


0.249 


66370 


0.51 


0.250 


67420 


0.52 


0.248 


67750 


0.52 


0.252 


61910 


1.03 


0.247 


67090 


1.02 


0.250 


66390 


1.02 


0.246 


66770 


1.02 


0.250 


67730 


1. 01 


0.247 


66020 


1. 00 


0.251 


67010 


1. 00 


0.247 


64450 


1. 01 


0.250 


66090 


1-54 


0.250 


64390 


1-52 


0.251 


63350 


1.50 


0.253 


64370 


1-54 


0.248 


64895 


2.02 


0.252 


64320 


2.00 


0.251 


62970 


2.00 


0.251 


60910 


2.50 


0.248 


59260 


2.50 


0.252 


63250 


2-53 


0.248 


5939o 


3-03 


0.251 


61577 


3.00 


0.249 


59080 


3.02 


0.251 


5955o 


3.02 


0.250 


59700 


3.00 


250 


63370 


3.00 


0.251 


58630 


3-03 


0.252 


63940 



RIVETED JOINTS. 



481 



Iron Punched. 



Iron Drilled. 



Steel Punched. 



Steel Drilled. 



Tensile Tests of 

Grooved Wrought- 

Iron Plates. 


1 


I 

.2 
h 


■fl.0 

C c 
>— < 

™ & 

lb 

P 


Inch. 


Inch. 




1. 01 


o-373 


47000 


0.98 


0.370 


47520 


2.00 


0.382 


39760 


2.02 


0.383 


36630 


2.39 


0.390 


37600 


2.98 


o-395 


36340 


2.98 


0.392 


39210 


3-47 


0.390 


37680 


3-47 


0.389 


38340 


0.97 


0.467 


50820 


1.48 


0.506 


45090 


1.49 


0.506 


45050 


1.91 


0.513 


42500 


1.97 


0.512 


43430 


2.47 


0.516 


394™ 


2.41 


o-5i3 


39720 


3.00 


0.515 


38950 


2.90 


0.517 


37290 


3-5° 


0.520 


37800 


3-49 


0.513 


37770 


4.00 


0.515 


35730 


4-o3 


0.516 


36690 


3-99 


0.511 


37000 


4-°3 


0.508 


37420 


0.97 


0.614 


49770 


1. 01 


0.619 


52960 


1.48 


0.618 


46320 


i-5 2 


0.620 


40750 


2.99 


0.614 


40140 


3-5° 


0.615 


3748o 


3-5° 


0.616 


36940 


4.04 


0.619 


373io 


0.98 


0.678 


50840 


1. 01 


0.682 


46590 


1.49 


0.688 


45970 


3-48 


0.691 


40350 


3-53 


0.692 


39380 



Tensile Tests of 

Grooved Wrought- 

Iron Plates. 


J2 


c 


c s 

p 


Inch. 


Inch 




0.98 


0.376 


50870 


0.98 


0-377 


52660 


1.98 


0-379 


49710 


2.00 


0.380 


49830 


2.50 


0.390 


50250 


3-oo 


0.392 


45150 


3.00 


o.393 


47540 


3-50 


0.392 


43940 


3-49 


0.390 


46490 


0.99 


o.477 


47140 


1. 00 


0.479 


48370 


1.49 


0.510 


51240 


1.49 


0.512 


51510 


1.98 


0.514 


50050 


1.98 


0.516 


47790 


2.51 


0.520 


45580 


2.52 


0.516 


44960 


3.00 


0.515 


44980 


3.01 


0.519 


47030 


3-5i 


0.513 


46170 


3-49 


0.514 


44760 


3-99 


0.510 


45330 


3.98 


0.513 


45000 


4.00 


0.506 


46100 


o.97 


0.628 


47220 


1. 00 


0.626 


48350 


152 


0.625 


47170 


1.49 


0.629 


46530 


2.98 


613 


48220 


3-46 


0.616 


47770 


3-47 


0.617 


44900 


3-9i 


0.625 


44840 


3-96 


0.626 


45100 


0.99 


0.695 


47500 


0.99 


0.691 


52780 


IS* 


0.692 


48470 


3-44 


0.700 


47750 


3-49 


0.692 


46350 



Tensile Tests 
of 

Grooved Steel Plates. 




g 

J* 
H 


■Sjs 
« 6- 

n £ 

is, 

P 


Inch. 


Inch. 




1.99 


0.365 


61890 


0.99 


o.494 


70080 


1. 00 


0.492 


68130 


1.50 


0.497 


66340 


1.51 


0.494 


63810 


1.99 


0.499 


55930 


1.97 


0.500 


64260 


2-43 


0.502 


52050 


2.51 


0.504 


64360 


3.00 


0-503 


60320 


2.99 


0.503 


62430 


3-5o 


0.503 


49430 


3-5o 


0.505 


48270 


4.00 


0.497 


48010 


4.00 


0.499 


55i9o 


3-99 


0.501 


5578o 


3-99 


0.498 


46250 


1. 01 


0.613 


66720 


1-52 


0.612 


64800 


i-5° 


0.615 


64400 


2.50 


0.618 


58060 


2.52 


0.619 


58780 


2.99 


0.617 


57180 


3.46 


0.615 


58410 


3-5i 


0.615 


57190 


4.04 


0.612 


5445o 


4-03 


0.614 


5738o 


1. 01 


0.721 


67930 


1. 00 


0.718 


67620 


1.50 


0.719 


62890 


3-50 


o.735 


56730 


3-5i 


0-733 


54220 



Tensile Tests 

of 

Grooved Steel Plates. 




10 

H 


•BjS 

C C 
1)1— 1 

™ & 

P 


Inch. 


Inch. 




1.97 


0.369 


63620 


1. 00 


0.498 


66220 


0.99 


0-495 


66800 


1. 00 


0.500 


67000 


i-53 


0-497 


65930 


1.50 


0.498 


66270 


1.98 


0.504 


67510 


2.03 


0.502 


66730 


2.50 


0.497 


67950 


2.52 


0.501 


67440 


3.01 


0.502 


66310 


3.01 


0.503 


66190 


3-49 


0.504 


64920 


3-50 


0.502 


65210 


3-99 


0.499 


64470 


4.00 


0.498 


64810 


4.00 


0.503 


64690 


4.00 


0.498 


64140 


099 


0.619 


60290 


1.49 


0.614 


63610 


1.49 


0.616 


63450 


2.49 


0.620 


59*70 


2.50 


0.619 


59600 


3.01 


0.617 


59270 


3-50 


0.614 


61610 


3-49 


0.617 


62060 


4.00 


0.615 


60330 


4.01 


0.617 


61 1 20 


0.96 


0.726 


58480 


1. 01 


0.727 


58790 


I-5I 


0.726 


59290 


3-50 


0.736 


58700 


3-49 


0.729 


59180 



482 APPLIED MECHANICS. 

§ 233. Chain Cable. — The most thorough set of tests of the 
strength of chain cable is that made by Commander Beardslee 
for the United-States government, an account of which may be 
found either in the report already referred to, or in the abridg- 
ment by William Kent. 

In this report are to be found a number of conclusions, 
some of which are as follows : — 

i°. That cables made of studded links (i.e., links with a 
cast-iron stud, to keep the sides apart) are weaker than open- 
link cables. 

2 . That the welding of the links is a source of weakness ; 
the amount of loss of strength from this cause being a very 
uncertain quantity, depending partly on the suitability of the 
iron for welding, and partly on the skill of the chain-welder. 

3 . That an iron which has a high tensile strength does not 
necessarily make a good iron for cables. Of the irons tested, 
those that made the strongest cables were irons with about 
51000 lbs. tensile strength. 

4 . The greatest strength possible to realize in a cable per 
square inch of the bar from which it is made being 200 per 
cent of that of the bar-iron from which it was made, the cables 
tested varied from 155 to 185 per cent of that of the bar- 
iron. 

5 . The Admiralty rule for proving chain cables, by which 
they are subjected to a load in excess of their elastic limit, 
is objected to, as liable to injure the cable: and the report 
suggests, in its place, a lower set of proving-strengths, as given 
in the following table ; the Admiralty proving-strengths being 
also given in the table. 

In these recommendations, account is taken of the different 
proportion of strength of different size bars as they come from 
the rolls, also no proving-stress is recommended greater than 
50 per cent of the strength of the weakest link, and 45.$ per 
cent of the strongest; whereas in the Admiralty tests, 66.2 



IRON AND STEEL WIRE. 



483 



per cent of the strength of the weakest, and 60.3 per cent of 
the strongest, is sometimes used. 

For the details of this investigation, see the report, Execu- 
tive Document No. 98, 45th Congress, second session, or the 
abridgment already referred to. 



Diameter of 

Iron, 

in inches. 


Recommended 
Proving-Strains . 


Admiralty 
Proving-Strains. 


Diameter of 

Iron, 

in inches. 


Recommended 
Proving- S trains . 


Admiralty 
Proving-Strains. 


2 


I21737 


161280 


I'fr 


66138 


833*7 


lit 


I 14806 


I5I357 


If 


60920 


76230 


T 7 


10S058 


I4I75 


i-fV 


55903 


69457 


III 


IOI499 


I3 2 457 


I* 


51084 


63000 


If 


95128 


I23480 


ifV 


46468 


56857 


Itt 


88947 


II4817 


ii 


42053 


5IO30 


If 


82956 


T0647O 


i* 


37820 


455 x 7 


T 9 


77159 


98437 


1 


33 8 4Q 


40320 


l\ 


7IS50 


9072O 









§ 234. Iron and Steel Wire. — It has long been known that 
the process of cold-drawing, by which wrought-iron and steel are 
made into wire, greatly increases its strength per square inch ; 
and usually the smaller the wire, the greater the strength per 
square inch. Thus iron wire varies in strength from about 70000 
to about 90000 lbs. per square inch when unannealed, its strength 
being reduced to 45000 or 50000 lbs. by being annealed. 

In steel wire unannealed, the strength runs up even to 
200000 lbs. per square inch in some cases ; and Fairbairn even 
records strengths as great as 275000 lbs. per square inch. 

Accounts of experiments upon the strength and elasticity 
of wrought-iron and steel wire will be found in — 

W. E. Woodbridge : Report on the Mechanical Properties of Steel. 
Professor Thurston : Materials of Engineering and other papers. 
Pocket-Book of the New-Jersey Iron and Steel Company. 



4§4 



APPLIED MECHANICS. 



The strength of wire becomes of special importance in 
connection with wire rope, and also with wire-wound guns. 
Dr. Woodbridge has made a great number of tests of steel wire 
for wire-wound guns. 

Wire Rope. — Inasmuch as wire rope is extensively used in 
the transmission of power, it may be a matter of convenience 
to have here a table giving the strength of the rope as claimed 
by sOme of the makers. There will follow, therefore, the table 
of strengths of the different sizes as given by Mr. John A. 
Roebling for the rope manufactured by him, and also some of 
his remarks in regard to its use : — 



Rope of 133 Wires, or 19 Wires to the 
Strand. 



Rope of 49 Wires, or 7 Wires to the 
Strand. 





■a 


,g 


•S.S 






bO 


£ -e 


& 


o 




8 S & 


g 


c 


c/5 


c S 


3 






a « w 


£ 


>a i/i 


«J </> 


«" O" « 


■8 


M 




Hi 


a 


y .5 


•j3 .c 


c W w 


H 


o 


53 


U 


I 


6f 


148000 


i5i 


2 


6 


130000 


i4i 


3 


5* 


108000 


13 


4 


5 


87200 


12 


5 


4l 


70000 


10! 


6 


4 


54400 


9\ 


7 


3t 


40400 


8 


8 


3i 


32000 


7 


9 


2| 


22800 


6 


IO 


2i 


17280 


5 


roi 


2 


10260 


A\ 


io* 


I 8 


8540 


4 


iof 


li 


6960 


3t 





.5 


j=" 


"SJ 






M 


V. -C 


u 





« 


So 

8 « E 


g 


c 




c a 


3 






ZMm 




<£J i/i 

II 

" .5 


1> (fl 


rcumfe 
Hemp 
Same 


H 





13 


U 


II 


4l 


72000 


I0| 


12 


4i 


60OOO 


IO 


13 


3f 


5OOOO 


9k 


14 


3t 


4OOOO 


8i 


IS 


3 


32O0O 


7\ 


16 


2f 


2460O 


61 


17 


at 


1660O 


$ 


18 


a* 


I520O 


5 


19 


1* 


II60O 


4l 


20 


if 


. 8l8o 


4 


21 


if 


5660 


3i 


22 


it 


4260 


2l 


2 3 


1* 


3300 


*i 


24 


1 


2760 


2i 


25 


1 


206o 


2 


26 


5 

4 


1620 


*4 


27 


I 


1 120 


I* 



IRON AND STEEL WIRE. 485 

Notes by Mr. Roebling. — "Two kinds of wire rope are 
manufactured. The most pliable variety contains nineteen 
wires in the strand, and is generally used for hoisting and 
running rope. The ropes with twelve wires and seven wires 
in the strand are stiffer, and are better adapted for standing- 
rope, guys, and rigging. Ropes are made up to three inches in 
diameter, both of iron and steel, upon special application. 

" For safe working-load, allow one-fifth to one-seventh of the 
ultimate strength, according to speed, so as to get good wear 
from the rope. When substituting wire rope for hemp rope, it 
is good economy to allow for the former the same weight per 
foot which experience has approved for the latter. 

"Wire rope is as pliable as new hemp rope of the same 
strength : the former will therefore run over the same size 
sheaves and pulleys as the latter. But the greater the diameter 
of the sheaves, pulleys, or drums, the longer wire rope will 
last. In the construction of machinery for wire rope, it will be 
found good economy to make the drums and sheaves as large 
as possible. 

" Experience has demonstrated that the wear increases with 
the speed. It is therefore better to increase the load than the 
speed. 

" Wire rope is manufactured either with a wire or a hemp 
centre. The latter is more pliable than the former, and will 
wear better where there is short bending. 

" Wire rope must not be coiled or uncoiled like hemp rope. 
When mounted on a reel, the latter should be mounted on a 
spindle or flat turn-table, to pay off the rope. When forwarded 
in a small coil without reel, roll it over the ground like a wheel, 
and run off the rope in that way. All untwisting or kinking 
must be avoided." 

In the case of wire rope it is true, as in all other cases, the 
only way to secure certainty in regard to the strength is to test 
it. The author has tested wire rope which bore no more than 



486 APPLIED MECHANICS. 

two-thirds the strength claimed for it ; though it is but justice 
to say, in this connection, that the few samples of Roebling 
rope tested in his laboratory bore out very fairly the results 
given for the sizes tested in the table. 

A number of tests of wire rope have been made on the 
government testing-machine at Watertown Arsenal and else- 
where. 

Wire rope is generally used with some kind of holder, and 
it is generally the case that the breakage of the rope occurs at 
the holder. 

Another matter worth noticing is, that it yields in detail, 
and hence that new rope is not as strong as that which has 
been under tension for some time, provided the tension has not 
been excessive, 

§235. Other Metals and Alloys. — Copper is, next to 
iron and steel, the metal most used in construction, sometimes 
in the pure state, especially in the form of sheets or wire, but 
more frequently alloyed with tin or zinc ; those metals where 
the tin predominates over the zinc being called bronze, and 
those where zinc predominates over tin, brass. Copper in the 
pure state was used not long ago for the fire-box plates of loco- 
motive and other steam-boilers, as it was believed to stand better 
the great strains due to the changes of temperature that come 
upon these plates, than iron or steel ; but now steel or iron has 
almost entirely superseded it for this purpose, except in some 
cases where the feed-water is very impure, and where the 
impurities are such as corrode iron. 

The alloys of copper, tin, and zinc which are used most 
where strength and toughness are needed, are those where the 
tin predominates over the zinc ; and the composition, mode of 
manufacture, and resisting properties of these metals, together 
with the effect of other ingredients, as phosphorus, have been 
very extensively investigated with reference to their use as a 
material for making guns ; instead of cast-iron. 



OTHER METALS AND ALLOYS. 



487 



Accounts of tests made on these alloys will be found as 
follows : — 

Major Wade: Ordnance Report, 1856. 

T. J. Rodman : Experiments on Metals for Cannon. 

Executive Document No. 23, 46th Congress, 2d session. 

No attempt will be made to give a complete account of the 
results of these tests ; but a table will be given for convenience 
of use, showing rough average values of the resisting powers of 
some metals and alloys other than iron. 





Specific 
Gravity. 


Tensile 
Strength 
per Sq. In. 


Modulus 

of 
Elasticity. 


Brass cast 


8.396 


1800O 


9170OOO 
I4230OOO 


Brass wire 


49OOO 


Bronze un wrought : 




84.29 copper + 1 5.7 1 tin (gun metal) 


8.561 


36060 


- 


82.81 " + 17.19 " 


8.462 


34048 


- 


81.10 " + 18.90 " 


8-459 


39648 


- 


78.97 " +21.03 " (brasses). . . 


8.728 


30464 


j 


34.92 " + 65.08 " (small bells) . 


8.056 


3^6 


- 


15.17 " +84.83 " (speculum metal) 


7447 


6944 


- 


Tin 


7.291 


5600 


- 


Zinc 


6.861 


7500 
24138 
33000 
60000 


_ 


Copper cast 


8.712 
8.878 




Copper bolts 




Copper wire 


I7O0OOOO 


Gold cast 


19.258 


20000 


Silver cast 


10.476 
22.069 
11.352 


40000 

56000 

1800 




Platinum wire 


- 


Lead cast 





Professor Thurston gives, for the ultimate tensile strength 
per square inch of a compound of copper and tin and zinc, 

T == 30000 +- 1000/ + 5002, 



488 APPLIED MECHANICS. 

where / = percentage of tin, and not over 15 per cent; and 
z = percentage of zinc, and not over 50 per cent. 

Some specimens of phosphor bronze tested by Kirkaldy 
gave for tensile strengths from 22000 to 50000 for cast, and 
98000 to 159000 for phosphor bronze wire 0.06 inch to o. 11 
inch diameter. If the student is to use alloys, he should ascer- 
tain their strength, elastic limit, and modulus of elasticity, and 
he should observe that the mode of manufacture has a great 
influence on the strength and ductility of any alloy. 

§ 236. Timber. — However extensively iron and steel may 
have superseded timber in construction, nevertheless, there are 
many cases in which iron is entirely unsuitable, and where 
timber is the only material that will answer the purpose ; and 
in many cases where either can be used, timber is much the 
cheaper. Hence it follows that the use of timber in construc- 
tion is even now, and as it seems always will be, a very impor- 
tant item. 

Another advantage possessed by timber is, that, on yielding, 
it gives more warning than iron, thus affording an opportunity 
to foresee and to prevent accident. 

If we make a section across any of the exogenous trees, as 
the oak, pine, etc., we shall find a series of concentric layers ; 
these layers being called annual rings, because one is generally 
deposited every year. 

Radiating from the heart outwards will be found a series of 
radial layers, these being known as the medullary rays. 

Of the annual rings, the outer ones are softer and lighter in 
color than the inner ones ; the former forming the sap-wood, and 
the latter the heart-wood. Wlien the log dries, and thus tends 
to contract, it will be found that scarcely any contraction takes 
place in the medullary rays ; but it must take place along 
the line of least resistance, viz., along the annual rings, thus 
causing radiating cracks, and drawing the rays nearer together 
on the side away from the crack. This action is exhibited in 



STRENGTH OF TIMBER. 



489 



Fig. 241, where a log is shown with two saw-cuts at right 
angles to each other ; when this log becomes dry, the four 



ris:ht 



through 



angles 



all 



becoming acute 



the 



shrinkage 



of the 




Fig. 241. 



rings. 

If the log be cut into planks by 
parallel saw-cuts, the planks will, 
after drying, assume the forms 
shown in Fig. 242, as is pointed 
out in Anderson's " Strength of 
Materials," from which these two 
cuts are taken. 

This internal construction of a 
plank has an important influence 
upon the side which should be uppermost when it is used for 

flooring ; for, if the heart side is up- 
permost, there will be a liability to 
having layers peel off as the wood 
dries : indeed, boards for flooring 
should be so cut as to have the an- 
nual rings at right angles to the 
side of the plank. Before discuss- 
ing any other considerations which 
affect the adaptability of timber to 
use in construction, we will con- 
sider the question of its strength. 
§ 237. Strength of Timber. — In this regard we must 
observe, that, whereas the strength and elasticity and other 
properties of iron and steel vary greatly with its chemical com- 
position and the treatment it has received during its manufac- 
ture, the strength, etc., of timber is much more variable, being 
seriously affected by the soil, climate, and other accidents of its 
growth, its seasoning, and other circumstances ; and that over 
many of these things we have no control : hence we must not 




490 APPLIED MECHANICS. 

expect to find that all timber that goes by one name has the 
same strength, and we shall find a much greater variation and 
irregularity in timber than in iron. The experiments that have 
been made on strength and elasticity of timber may be divided 
into the following classes : — 

1°. Those of the older experimenters, except those made 
on full-size columns by P. S. Girard, and published in 1798. 
A fair representation of the results obtained by them, all of 
which were deduced from experiments on small pieces, is to 
be found in the tables given in Professor Rankine's books, 
"Applied Mechanics," " Civil Engineering," and "Machinery 
and Millwork." 

2°. Tests made by modern experimenters on small pieces. 
Such tests have been made by — 

(a) Trautwine : Engineers' Pocket-Book. 

(<£) Hatfield : Transverse Strains. 

(<:) Laslett : Timber and Timber Trees. 

(//) Thurston : Materials of Construction. 

(e) A series of tests on small samples of a great variety of American 
woods, made for the Census Department, and recorded in 
Executive Document No. 5,48th Congress, 1st session. 

3 . Tests made by Capt. T. J. Rodman, U.S.A., the results 
of which are given in the "Ordnance Manual." 

4 . All tests that have been made on full-size pieces. 

In regard to tests on small pieces, such as have commonly 
been used for testing, it is to be observed, that, while a great 
deal of interesting information may be derived' from such tests 
as to some of the properties of the timber tested, nevertheless, 
such specimens do not furnish us with results which it is safe 
to use in practical cases where full-size pieces are used. Inas- 
much as these small pieces are necessarily much more perfect 
(otherwise they would not be considered fit for testing), having 
less defects, such as knots, shakes, etc., than the full-size pieces, 



STRENGTH OF TIMBER. 



491 



they have also a far greater homogeneity. They also season 
much more quickly and uniformly than full-size pieces. In 
making this statement, I am only urging the importance of 
adopting in this experimental work the same principle that the 
physicist recognizes in all his work ; viz., that he must not 
apply the results to cases where the conditions are essentially 
different from those he has tested. 

Moreover, it will be seen in what follows, that, whenever 
full-size pieces have been tested, they have fallen far short of 
the strength that has been attributed to them when the basis 
in computing their strength has been tests on small pieces ; 
and, moreover, the irregularities do not bear the same propor- 
tion in all cases, but need to be taken account of. 

The results of the first class of experiments named in the 
following table are taken from Rankine's "Applied Mechanics;" 
and, inasmuch as the table contains also the strengths of some 
other organic fibres, it will be inserted in full. The student 
may compare these constants with those that will be given 
later. 



Kind of Material. 


Tenacity 
or Resist- 
ance to 
Tearing. 


Modulus of 

Tensile 
Elasticity. 


Resist- 
ance to 
Crush- 
ing. 


Modulus 

of 
Rupture. 


Resist- 
ance to 
Shearing 
along 
Grain. 


Modulus 

of 

Shearing 

Elasticity 

along the 

Grain. 


Ash 

Bamboo 

Beech 

Birch 

Blue gum 

Box 

Bullet-tree .... 
Cedar of Lebanon . . 


I70OO 

63OO 

1 1 500 

150OO 

20000 
1 1400 


160OOOO 

I350OOO 
16450OO 

486000 


9000 

9360 

640O 

8800 

IO300 

' I4000 

5860 


( 12000 
( I40OO 

j 9000 
\ 12000 
1 1 700 
( 160OO 
( 20000 

( I59OO 

1 160OO 

7400 


I40O 


76000 



492 



APPLIED MECHANICS. 



Kind of Material. 



Tenacity 
or Resist- 
ance to 
Tearing. 



Modulus of 

Tensile 
Elasticity. 



Resist- 
ance to 
Crush- 
ing. 



Modulus 

of 
Rupture. 



Resist- 
ance to 
Shearing 
along 
Grain. 



Modulus 

of 

Shearing 

Elasticity 

along the 

Grain. 



Chestnut 
Cowrie . 



Ebony 



Elm 



Fir, Red pine . . 
" Yellow pine (Am 
" Spruce . . 

" Larch . . 

Hoxen yarn . . 
Hazel .... 



Hempen rope 

Ox-hide, undressed 
Hornbeam . . 
Lancewood . . 
Ox-leather . . 
Lignum-vitae . . 
Locust .... 



Mahogany . . 

Maple . . . 

Oak, British . 

" Dantzic 



European . 
American red 



ioooo 

to 
13000 



14000 

12000 

to 
14000 



12400 

9000 

to 

1 0000 

25000 

16000 

I20OO 

to 

IOOOO 

63OO 

2OO00 

234OO 

4200 

1 1800 

l6000 

80OO 

to 
21800 
10600 



IOOOO 

to 
19800 
10250 



I I 40000 



( 700000 
] to 
( 1840000 

1460000 
to 

1900000 




24300 



155000 



1 200000 

1750000 

2 I 5OOOO 



19000 
IO3OO 

5375 

to 

6200 

5400 



5570 



9900 



IOOOO 

7700 

6000 



10660 

1 1 000 

27000 

6000 

to 

< 9700 

7100 

9540 



9900 



5000 

IOOOO 



7600 
1 1 500 



IOOOO 

13600 
8700 



10600 



1400 



500 

800 



600 



970 

1700 



2300 



76000 



62000 
1 1 6000 



82000 



STRENGTH OF TIMBER. 



493 



Kind of Material. 


Tenacity 
or Resist- 
ance to 
Tearing. 


Modulus of 

Tensile 
Elasticity. 


Resist- 
ance to 
Crush- 
ing. 


Modulus 

of 
Rupture. 


Resist- 
ance to 
Shearing 
along 
Grain. 


Modulus 

of 

Shearing 

Elasticity 

along the 

Grain. 




52000 
I3000 

15000 

21000 
77OO 

8000 


130OOOO 
I040000 

2400OOO 

2300000 


12000 


9600 
( 12000 
( 190OO 

14980 
6600 


\\ 


- 






Teak, Indian 

" African 
Whalebone . 
Willow . . . 











In regard to the tests of the second class, a few comments 
are in order : — 

i°. These experiments, like those of the first class, were all 
made upon small pieces ; and the results are correspondingly- 
high. 

The usual size of the specimens for crushing being one or 
two square inches in section, and of those for transverse 
strength being about two inches square in section and four or 
five feet span, those for tension had even a much smaller sec- 
tion than those for compression ; as it is necessary, in order to 
hold the wood in the machine, to give it very large shoulders. 

The only exception to this is the tests of Sir Thomas Las- 
lett, an account of which is given in his " Timber and Timber 
Trees," and also in D. K. Clark's " Rules and Tables." In these 
tests he gives very much lower tensile strengths than those 
given above ; and he states that his specimens were three inches 
square, but does not say how he managed to hold them in such 
a way as to subject them to a direct tensile stress. His results 
for crushing and transverse strength are about as great as 



494 APPLIED MECHANICS. 

those given in Rankine's tables, and as were obtained by the 
other experimenters on small pieces, as his specimens were of 
about the same dimensions as those used by the others. The 
figures obtained by these experimenters will only be given inci- 
dentally, as — 

{a) They are very similar to those given in Rankine's table. 

(b) They are not suitable for practical use on the large 
scale. 

(c) While they have been used, it has only been done by 
employing a very large factor of safety for timber. 

The series of tests made for the Census Department, and 
recorded in Executive Document No. 5, 48th Congress, first 
session, form a very interesting series of experiments upon 
small specimens of an exceedingly large number of American 
woods. In order to have working figures, we should need to 
test large pieces of the same ; as the proportion between the 
strengths of the different kinds would be liable to be different 
in the latter case. 

We will next consider Rodman's experiments. The only 
record of them available is a table of results in the " Ordnance 
Manual," and this table is appended here. It will be seen, on 
comparing it with the former table, that the results are lower 
as a rule than those obtained by the experimenters of the first 
or second class. This is to be accounted for by the fact, that, 
while he did not experiment on full-size pieces, he used much 
larger pieces than those heretofore employed ; his specimens 
for transverse strength, many of which are still stored at the 
Watertown Arsenal, being 5|- inches deep, 2J inches thick, 
and 5 feet span. 

In the table, the modulus of rupture will be given, instead 
of one-sixth of its value, which is the constant given in the 
" Ordnance Manual." 



STRENGTH OF TIMBER. 



495 



Material. 


Locality. 


Length 

of 
Season- 
ing. 


Crushing 

Force per 

Square 

Jach. 


Tensile 
Strength 

per 
Square 
Inch. 


Modulus 
of Rup- 
ture, in 
lbs., per 
Square 
Inch. 






years. 


lbs. 


lbs. 




Ash 


Ohio 


15 


8783 


24033 


12708 


" . 










Pennsylvania 




3 


4475 


14266 


8796 


(< 










Canada . . 




9 


557i 


15000 


- 


" . 










New York 






7 


4783 


II786 


- 


«« 










Vermont 






2 


5858 


10893 


15984 


«( 










Virginia 
Oregon . 






1 

1 


6663 
5789 


23167 

14700 


9168 
8628 


Birch 










Maine . 






4 


7969 


r 5333 


16776 


Bass . 










« 
Canada . 






12 
9 


5 2 7i 
4609 


12600 
H953 


1 1 478 


Box . 










Africa . 






5 


105*3 


23600 


- 


Fir, White 






Oregon . 






2 


6644 


J 4533 


4194 


Gum, Black 






Alabama 






1 


6703 


15860 


8886 


Hickory 






Ohio . . 






13 


9887 


25900 


16362 


" 






North Carolina . . 


3 


6125 


18000 


- 


" 






Eastern Virginia 


1 


5492 


355oo 


- 


Red 




Massachusetts . . 


7 


10942 


27133 


17400 


« << 




New York. . . . 


7 


7725 


12866 


16536 


White 




Massachusetts . . 


7 


8925 


38700 


17316 


U <( 




Alabama . . . . 


1 


11213 


40067 


1 68 18 


« M 




Virginia 




1 


9733 


36666 


20352 


Holly . . . 




" 


. . . 


1 


5246 


18567 


3384 


Hemlock . 






Oregon . 


. . . 


1 


6817 


16533 


7752 


Hackmatack 






Maine . 




[ 


- 


- 


7860 


Lignum-vitse 






South America . . 


4 


9854 


16000 


16080 


Locust . . 






Pennsylvania . . . 


1 


9113 


275 T 7 


14478 


Mahogany 








San Domingo . . 


4 


7390 


12350 


9996 


Maple . 








Canada 


9 


7716 


22933 


- 


<< 








Maine 


4 


8621 


21720 


1 1 574 


«< 








Oregon 


1 


4443 


10400 


5838 


Oak, White 






New England . . 


18 


6668 


19600 


10980 


<( << 






Western New Jersey, 


12 • 


6620 


19166 


11256 


<< « 






Ohio 


T 3 


6258 


19066 


8754 


H M 






Monongahela . . 


13 


6592 


20333 


12216 



496 



APPLIED MECHANICS. 













Modulus 


Material. 


Locality. 


Length 

of 
Season- 


Crushing 

Force per 

Square 


Tensile 
Strength 

per 
Square 
Inch. 


of Rup- 
ture, in 
lbs., per 






ing. 


Inch. 


Square 












Inch. 






years. 


lbs. 


lbs. 




Oak, White . . . 


Ohio 


5 


9108 


19466 


17340 




■ " 






New York .... 


ii 


4691 


12300 


10668 




« « 






Maryland . . . . 


19 


6092 


17666 


14556 




< <« 






Massachusetts . . 


43 


5800 


16766 


I4658 




< u 






<( 


7 


7292 


19200 


1 1 700 




1 " 






" (pasture), 


7 


6962 


16200 


J3596 




( (( 






Canada 


9 


60OO 


16646 


- 




( «( 






Connecticut . . . 


14 


5*99 


^3333 


- 




< «« 






" ... 


18 


7089 


21000 


- 




« (( 






North Carolina . . 


8 


6550 


21 100 


- 




« (( 






Alabama . . . . 


2 


5744 


18307 


9912 




( (« 






Virginia .... 


1 


6902 


19033 


10758 




« H 






Oregon 


1 


6072 


18467 


943 2 




( (( 






James River, Va. . 


13 


6667 


2 ^222 


10938 




■ Yellow 






New Hampshire 


13 


6279 


25000 


1 1 490 




1 Live . 






Alabama .... 


3 


6531 


163S3 


9780 




1 " 






- 


- 


7279 


15800 


7998 


Pi 


ne, Pitch 






North Carolina . . 


3 


8947 


1 1400 


- 




4 White 






Alleghany River 


4 


5 OI 7 


"433 


6798 




1 " 






New York .... 


5 


5775 


1 1 933 


6912 




' " 






Maine 


13 


5 6l 7 


1 1960 


7092 




' Yellow 






Florida 


6 


8350 


18000 


8796 




« «« 






North Carolina . . 
Alabama . . ■ . 


, 


7836 
8201 


12600 
17946 


1 1676 
10254 




« « 






Virginia .... 


2 


7867 


19200 


9168 




1 Sugar 






Nevada Co., Cal. . 


1 


- 


- 


53 2 2 




« <« 






Humboldt Co., Cal., 


1 


- - 


- 


5658 


Poplar . . 






Ohio 


3 


5742 


14933 


7260 


«« 






New York . . . 


2 


6075 


9066 


5856 


M 






Virginia .... 


1 


6579 


8200 


7782 


Redwood . 






California . . . . 


1 


6083 


10833 


4518 


Spruce . . 






Maine 


1 


6862 


13666 


6168 


" . . 






Oregon 


1 


5092 


10867 


5964 


Teak . . 






East India .... 


4 


10819 


33800 


18558 



TENSION. 



497 



Material. 


Locality. 


Length 

of 
Season- 
ing. 


Crushing 

Force per 

Square 

Inch. 


Tensile 
Strength 

per 
Square 
Inch. 


Modulus 
of Rup- 
ture, in 
lbs., per 
Square 
Inch. 


Walnut, Black . . 
« « 


Western States . . 
Virginia .... 
Michigan .... 
Canada 


years. 

7 

i 

2 

9 


lbs. 

7471 

7500 

5782 

5989 


lbs. 
16633 
163OO 
I7580 
16133 


I2 3 l8 

8190 



The fourth class of tests are those which furnish reliable 
data for use in construction ; and we will proceed to a consid- 
eration of these, taking up (i°) tension, (2 ) compression, (3 ) 
transverse strength, and (4 ) shearing along the grain. 



TENSION. 

In all cases where the attempt has been made to experiment': 
upon the tensile strength of timber, a great deal of difficulty 1 ' 
has been encountered in regard to the manner of holding the 
specimens. In all cases it has been found necessary to pro- 
vide them with shoulders, each shoulder being five or six times 
as long as the part of the specimen to be tested, and to bring 
upon these shoulders a powerful lateral pressure, to prevent: 
the specimen from giving way by shearing along the grain, andi 
pulling out from the shoulder, instead of tearing apart. 

The specimens tested have generally- had a sectional area 
less than one square inch, and it seems almost impossible to 
provide the means of holding larger specimens. This being 
the case, it is plain, that, whenever timber is used as a tie-bar 
in construction (except in exceedingly rare and out-of-the- 
way cases), it will give way by some means other than direct 



498 APPLIED MECHANICS. 

tension ; i.e., either by the pulling-out of the bolts or fastenings, 
and the consequent shearing of the timber, or else by bending 
if there is a transverse stress upon the piece ; and, this being 
the case, these other resistances should be computed, instead 
of the direct tension. Hence, while the direct tensile strength 
of timber may be an interesting subject of experiment, it can 
serve hardly any purpose in construction ; and the conclusion 
follows, that the resistances of timber to breaking we may 
expect to meet in practice are its crushing, transverse, and 
shearing strength. Indeed, the use of timber for a tie-bar 
should be avoided whenever it is possible to do so ; and, when 
it is used, the calculations for its strength should be based 
upon the pulling-out of the fastenings, the shearing or splitting 
of the wood, etc., and not on the tensile resistance of the solid 
piece. 

Moreover, when a wooden tie-bar is loaded, in addition, with 
a transverse load, and when the magnitude of this transverse 
load is so great as to render the piece more liable to give way 
by cross-breaking than by the pulling-out of the fastenings, we 
must then compute the greatest tension per square inch at the 
outside fibre due to the bending, and to that add the direct 
tension per square inch : and this sum must be less than the 
modulus of rupture if the piece is not to give way ; i.e., the 
modulus of rupture, and not the ultimate tensile strength per 
square inch, must be our criterion of breaking in such a case, 
the working-strength per square inch being the modulus of 
rupture divided by a suitable factor of safety. 

COMPRESSIVE STRENGTH. 

Tests of the compressive strength of full-size wooden col- 
umns are, with the exception of one set of tests, of very recent 
date, and have not yet found their way, to any extent, into our 
text-books and engineers' handbooks. It is therefore only 



COMPRESSIVE STRENGTH. 499 

suitable, that, before enumerating them, we should observe 
what formulae have been given in these books for the computa- 
tion of timber columns in practice, what experimental basis 
these formulae rest upon, and how they coincide with the 
facts. 

The oldest formulae are those of Euler. His formulae for 
the strength of such circular columns as yield by bending are 
as follows : — 

For ends fixed in direction, 



For rounded ends, 



(3.1416)3 ^4 



(3.1416)3 ^ 



where P = breaking- weight in pounds, and E = modulus of 
elasticity of the material. For wood, Weisbach gives for use 
in Euler's formulas, E = 1664000, and crushing-strength per 
square inch = 6770. 

According to Hodgkinson, we should take one-ninth of the 
breaking-strength of a cast-iron pillar, as given by equations. 
(1), (2), (3), (4), (5), § 211 ; this being the rule given by Mr. 
James B. Francis in his book, where he has computed and 
tabulated the strength of pillars of the ordinary sizes used in 
practice, by means of Mr. Hodgkinson's formulae. 

In Gordon's formula the constants used were determined 
in such a way as to make the results agree as nearly as possible 
with Hodgkinson's experiments. The formulae devised by Gor- 
don himself refer only to cylindrical and hollow cylindrical 
columns. A formula devised by the same course of reasoning, 
and also depending for its constants on Hodgkinson's experi- 
ments, but so arranged as to be applicable to any form of 



500 



APPLIED MECHANICS. 



section whatever, is given by Professor Rankine. For wood 
it is as follows : 

72006" 



P = 



1 + 



3Qoor 2 



where P = breaking-strength in pounds, 5 = sectional area in 
square inches, / = length in inches, r = least radius of gyra- 
tion in inches. 

Besides the above, we have formulae which are practically 
Rankine's formulae with the constants changed. One of these 
is that of Mr. C. Shaler Smith as given in Trautwine's " Pocket- 
Book," and applicable, as he claims, to a square or rectangular 
column of white or yellow pine. It is as follows : 



50006' 



/2 ' 
1 -+- 0.004^ 

d 2 



where P = breaking-weight in pounds, ^ = sectional area in 
square inches, /= length in inches, d ' = least side of rectangle. 

I have computed, by means of these formulae, the breaking- 
weights of certain oak columns, with the following results : — 

Length = 14 feet. 







Diameter 


Diameter 






10.5 Inches. 


9.5 Inches. 


Euler . . . 


( Flat ends 


1 586214 


479858 


^ Rounded ends . . . 


347M7 


232623 


Rankine . . 


( Flat ends 


465261 


360164 


1 Rounded ends . . . 


263615 


191286 


Francis . . . 


C Flat ends 


420OOO 


3 I2 795 


( Rounded ends . . . 


140000 


104265 



COMPRESSIVE STRENGTH. 501 

If we use the average of the results obtained from the oak 
columns tested at Watertown, which will be referred to later, 
we should find, for the breaking-weights of the above columns 
with flat ends, about 277000 and 227000 lbs. respectively. 

A glance at the above results will show that they differ very 
much from each other, and the question naturally arises as to 
the trustworthiness of the experimental data on which they 
are based. 

The constants used in Euler's formulae are not deduced 
from any experiment on the breaking of a column. 

Rankine's and Francis's have, for experimental basis, the 
experiments of Hodgkinson. He made a very large number 
of tests of cast-iron columns, none of which were as large as 
those used in practice. On oak columns, he made seventeen 
experiments on as many columns, all cut from one good plank 
of Dantzic oak, the largest of which was five feet long and two 
inches square. Of these seventeen, only seven were used in 
deducing his formulae. 

It is plain that such data are insufficient, and cannot furnish 
us reliable information as to the strength of a column. 

As to Mr. C. Shaler Smith's formula, it is based upon some 
experiments made by himself, which have never been published. 
The student can easily satisfy himself as to how accurately it 
represents the facts on the large scale, by computing by it the 
strength of some of the columns tested, of which an account 
will shortly be given. Trautwine, in his " Handbook,'* states 
that it is "for the breaking-loads of either square or rectangular 
pillars or posts, of moderately seasoned white and common 
yellow pine, with flat ends, firmly fixed, and equally loaded." 

Mr. Smith, as I understand, however, intended it to meet 
the case of such ill-fitting joints as occur in practice, and not 
for perfectly even bearings. 



502 APPLIED MECHANICS. 



TESTS OF FULL-SIZE COLUMNS. 

The only tests of full-size columns of which I have any 
knowledge are : — 

i°. Trautwine, in his "Handbook," speaks of some tests of 
wooden pillars 20 feet long and 13 inches square, made by 
David Kirkaldy, which, as he says, gave results agreeing with 
Mr. C. Shaler Smith's rule. 

2°. A series of tests made at the Watertown Arsenal for 
the Boston Manufacturers' Mutual Fire Insurance Company, 
under the direction of the author. 

3 . The tests that have been made at the Watertown Arse- 
nal on the government testing-machine. 

4 . Besides these, the writer has very recently had his 
attention called to a series of tests of full-size columns of oak 
and fir, made by P. S. Girard in 1798, which give results agree- 
ing very well with the modern results on full-size columns. 
Why these tests should have been lost sight of, and Hodgkin- 
son's always used instead, is incomprehensible. 

In regard to the first, no details or results are given : hence 
nothing will be said about them. 

In regard to the second, a summary only will be presented 
here, the reader being referred for details to my published 
report entitled " Strength of Wooden Columns." 



TESTS OF YELLOW-PINE POSTS AND BLOCKS. 



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TESTS OF OLD AND SEASONED WHITE-OAK POSTS. 505 





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APPLIED MECHANICS. 



In all the experiments enumerated in the tables given 
above, the columns gave way by direct crushing, and hence 
the strength of columns of these ratios of length to diameter 
can properly be found by multiplying the crushing-strength per 
square inch of the wood by the area of the section in square 
inches. 

This conclusion is deduced from the fact that the deflections 
were measured in every case, and found to be so small as not to 
exert any appreciable effect. 

In regard to other tests of this same set, there were eight 
tests made, in addition to those already enumerated ; and of 
these, five were loaded off centre. A summary of the results is 
appended, together with a comparison of their actual strength 
with that which would be computed on the basis of 4400 per 
square inch for yellow pine, and 3000 for oak. 





Weight, 
in lbs. 


Length, in 
feet and 
inches. 


Diameter 

of 
Column. 


Diam- 
eter of 
Core. 


Sectional 
Area, in 
square 
inches. 


Eccen- 
tricity, 

in 
inches. 


Ultimate 
Strength. 


Computed 
Ultimate 
Strength. 






ft. in. 














2, 2d series 


320 


II II.27 


9.92 


i-53 


7545 


2-33 


2650OO 


331980 


5, 3d series 


298 


12 6.8 


\ 7.60 ) 


- 


63.1 


2.07 


240OOO 


277640 


1, 3d series 


386 


12 9.3 


C 8.7s) 
I 8.92) 


- 


76.04 


2.25 


280000 


334576 


1, 2d series 


45 1 


II II.4 


10.95 


1.80 


92.16 


2-75 


170000 


276480 


3, 2d series 


236 


II II. 2 


8.2 


i-55 


50.92 


1. 91 


I OOOOO 


152760 



These results exhibit a great falling-off of strength due to 
the eccentricity of the load ; and if we observe, that, whenever 
the beam on one side of a column is loaded differently from 
that on the other side, we have an eccentric loading, and hence 



STRENGTH OF TIMBER. 507 

a falling-off in strength, we must conclude that this should be 
taken into account. Probably the best way to proceed in the 
matter is to compute always the greatest eccentricity possible 
in any given case, and to compute from that the additional 
stress on the column due to the bending consequent on this 
eccentricity by the principles of the short strut, already ex- 
plained in § 207. This will always be on the safe side. The 
three remaining experiments of the set were, (i°) Two tests of 
whitevvood columns : these gave a crushing-strength of 3009 
lbs. per square inch. The columns were, however, brittle, and 
did not give warning of fracture. (2 ) One yellow-pine square 
column of a sectional area of 68.8 square inches, and length 12 
feet 6.85 inches, tested by resting one end on a thick yellow- 
pine bolster, crushing this bolster at right angles to the grain. 

Maximum load on the post while the bolster was in, was 
120000 lbs. = 1744 lbs. per square inch. Under this load a 
crack at end of post was enlarged, giving evidence that failure 
of the post was imminent. The bolster in the mean time had 
become thoroughly cracked, owing to the unequal distribution of 
the load on the bolster, from imperfect workmanship. Slight 
cracks followed the first snapping sounds heard at 20000 lbs. 
compression. The cracks gradually developed as the loads were 
increased, the side nearest the heart of the bolster sliding off. 

The post was taken from machine and bolster removed. 
The post was cut off 1% inches at the end, and squared ; the 
total length, after cutting, being 149.35 inches. Ultimate 
strength, 375000 lbs. = 5451 lbs. per square inch. 

A few of the conclusions in regard to these sizes of posts will 
now be quoted from the report of these tests, in which some 
general recommendations are made, and others having special 
reference to mill columns : — 

1. I should recommend that the longitudinal holes in wooden mill 
columns be bored from one end only, and that all posts be rejected in 



508 APPLIED MECHANICS. 

which the eccentricity at the other end is greater than a given small 
amount, as three-quarters of an inch. This recommendation is made in 
view of the facts that holes bored from the two ends are very liable not 
to meet in the middle, and hence not to allow a circulation of air ; that, 
if the hole becomes very eccentric, the column is liable to be weakened ; 
and also by the presence of two holes at the same section. 

2. I should recommend that mill columns be not tapered, as the 
tapering is a source of weakness ; the loss of strength in one of the cases 
tested amounting to about 120000 lbs. 

3. I should also recommend that square columns be used in mills, 
instead of round ones, for the reason that the timber comes to the wharf 
in the form of square logs, and, when the columns are made round, they 
are cut from the square form ; and this cutting-away of the wood is so 
much loss of strength. 

4. The strength of a column of hard pine or oak, with " flat ends," 
the load being uniformly distributed over the ends, and of the diameters 
tested, is practically independent of the length up to a length of twelve 
feet (how much farther can only be decided by further experiment), such 
columns giving way practically by direct crushing ; the deflection, if any, 
being as a rule very small, and exerting no appreciable influence on the 
breaking-strength. 

5. The only exceptions to the above are found in cases where there 
is good reason for departure from the rule, as in the case of very imper- 
fect wood or of very eccentric holes ; but even there the influence of 
the deflection in reducing the strength is not nearly so great as has been 
generally supposed. 

6. No formulae founded on the generally received hypothesis, that 
the deflection exerts a very considerable influence on the breaking- 
strength of such columns as those referred to, represent correctly their 
breaking-strength for all lengths and diameters. 

7. For such columns as those referred to, the most correct rule for 
determining the breaking-strength is to multiply the number of square 
inches in the section (the smaller section being used in the case of 
tapering columns) by the crushing-strength per square inch of the 
wood. 

8. The crushing-strength per square inch varies considerably in 



STRENGTH OF TIMBER. 509 

specimens of different degrees of seasoning, also in large and small 
specimens. 

9. The average crushing-strength of wood is much less than has 
been supposed by many. That of some very highly seasoned hard pine 
was found at the arsenal to be 7386 lbs. For some hard pine of very 
slow growth and very highly seasoned, an average crushing-strength was 
found of 9339 lbs. For some very wet and green, they found a crush- 
ing-strength of 3015 lbs. For some yellow pine which had been season- 
ing about three months, I found 5400 lbs. For average crushing-strength 
of such posts as I tested, not thoroughly seasoned, and not very green, 
I found about 4400 lbs. ;. whereas in none of these cases did I obtain a 
greater result than about 4700 lbs. Hence it would be entirely unfair 
to assume a crushing-strength of 8000 lbs. for yellow pine. For two 
specimens of white oak tried at the arsenal, and very thoroughly sea- 
soned, an average was obtained of about 7150 lbs.; whereas for such 
oak as was furnished me, which was green and knotty, but no more 
so than is usual for use in building, I obtained an average of about 
3200 lbs. 

10. I would recommend the use of iron caps and pintles, instead of 
wooden bolsters, as wood is very weak to resist crushing across the 
grain, and the wooden bolster will fail at a pressure far below that which 
the column is capable of resisting ; and the unevenness of the pressure 
brought about by the bolster is so great as to sometimes crack the col- 
umn at a pressure far below what it would otherwise sustain. 

11. Any cause which operates to distribute the pressure on the ends 
unevenly, or to force its resultant out of centre, is a source of weakness, 
and brings about a very considerable deflection, which exerts an impor- 
tant influence in reducing the breaking-strength. 

12. As far as these experiments have gone, it appears that such pin- 
tles as were used in these tests, when the fitting is perfect, exert no 
influence upon the breaking-strength of perfect and straight-grained 
columns, but that they probably are a source of weakness in the case of 
imperfect and knotty wood, and especially in cases where there is an 
incipient deflection. Further experiment is needed, however, to answer 
these questions fully. 

13. I would also recommend that the horizontal holes 



5IO APPLIED MECHANICS. 

the longitudinal holes with the outside air, be made in the iron cap, and 
not in the wood : this will prevent weakening of the post by the hole, 
and will prevent the closing of the hole by change in moisture and 
other causes. 

14. Another conclusion which I think is very evident, is, that the 
crushing-strength of full-size columns cannot be fairly inferred from tests 
made on columns no larger than five feet long and two inches on a 
side. 

The table of results of the tests on old and seasoned oak 
columns were made upon columns that had been in use for a 
number of years in different mills, from which they were re- 
moved, and replaced by new ones. Ten of them had been in 
use about twenty-five years, and the remainder for shorter 
periods. An inspection of this table will, I think, convince the 
reader that it would not be safe to calculate upon a higher 
breaking-strength per square inch in these than in the green 
ones. 



TESTS MADE ON THE GOVERNMENT MACHINE. 

In Executive Document 12, 47th Congress, first session, 
will be found a series of tests of white and yellow pine posts 
made at the Watertown Arsenal ; and these tests probably fur- 
nish us the best information that we possess in regard to the 
strength of wooden columns. 

The summary of results is appended : — 



COMPRESSION OF WHITE-PINE POSTS. 



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o 


o 

VO 


o 

oo 


o 

n 


vS 


O 

vq 


o on 




t^. 


vd 


tA. 


vd 


r^. 


r-^. 


t>. 


co 


CO 


CO 


it 


4 


-5- 


4 


in 


in 


>-o 


in 


On 4 




D< 


N 


ci 


0) 


Cl 


M 


04 


04 


M 


M 


04 


r^ 


r^ 


r^ 


r~^ 


r-~x 


r^ 


r^ 


r^ 


vo r^ 




1) 




ON 


CO 


<N 


CO 


in 


in 


«n 


rh 


in 


in 


-f 


o 


OO 


01 


VO 


^i- 


in 


lO 


rn c-> 






0) 




N 




Cl 


01 


01 


rn 


rn 


OO 


vo 


r> 


VO 


vo 


r^ 


r^ 


i>» 


t^. 


04 "vO 




C 


i-O 


iO 


in 


IO 


>n 


LO 


in 


in 


in 


in 


On 


On 


ON 


c\ 


On 


cK 


ON 


ON 


On On 


J 


A 










































J3 




1^ 


vo 


O 


CO 


r^ 


in 


>n 


01 


in 


>n 


r-». 


O 


r^~) 


in 


01 


on 


in 


m oo n 


1 

o 


•^ 


C 


01 


M 


Cl 


M 


01 


N 


01 


on 


T 5 


rn 


r-. 


r^. 


t^. 


r^- 


l^ 


rj. 


r-. 


r^ 


T ^ 


£ 




in 


in 


LO 


UO 


>n 


in 


in 


in 


•n 


in 


rA 


r^> 


r^. 


r^ 


r^ 


t^. 


r^- 


r^. 


r^» t-^ 






CO 


VO 


ro VO 


H 


-t- 


r^ 


in 


o 


O 


r^ 





oo 


ro vo 


oo 


01 


■* 


Tf o 


w 


• 


a 


N 


N 


N 


N 


CO 


oo 


oo 


on 


Tf 


<n 


O 
















i-J <N 




C 
3 




d 


vo 


vd 


vd 


d 


d 


d 


vd 


vd 


vd 


CO 


CO 


CO 


d 


d 


d 


4 


4 


4 00 




. 


o 


N 


N 


M 


in 


in 


in 


r^ 


r^. 


0^ 


VO 


vO 


vo 


o 


o 


o 


OO 


oo 


m vo 






0) 


N 


Cl 


04 


01 


N 


N 


01 


0) 


01 






















t/> c 








































Avera 

Rat 

of 

Grow 


W- 


1 


1 


1 


| 


1 


1 


1 


1 


1 


1 


1 


r^ 


01 


in 


M 


o 


r^. 


m 


rn r^ 


>- p. 


























M 


" 


H 


M 






M 






q 


O 


q 


o 


q 


q 


q 


O 


O 


q 


O 


O 


in 


o 


o 


o 


q 


q 


in q 


u? 


»-0 


d 


ro od 


rs. 


in 


A 


^ 


on 


r^. 


4 


0O 


d 


vd 


vd 


vg 


— 


oi 


vo o4 


'5 


jQ 


Ox 






M 


01 


01 


01 


t~s 


in 


ON 


On 


O) 


vo 


vo 


01 


CO 


ON Tt 


£ 








M 


H 


M 


~ 


~ 


M 


















04 




H-c 04 


^<*« 




vo 


in vo 


t*s 


oi 


O 


t*- VO 


r> 


^N 


On 


O 


M 


in 


vo 


r-. 


_ 


04 


oo ON 


£ o.iu 




w 


o 


o 


o 


O 


O 


O 


o\ 


ON 


VO 


ts. 


r-^ 


r^ 


r^ 


r-~ 


CO 


CO 




t"< 




m 


UO 


iO 


LO 


m 


in 


in 


T}- 


«fr 


^J" 


in 


in 


in 


in 


>n 


in 


■n 


LO 


in in 1 



COMPRESSION OF WHITE-PINE POSTS. 



513 







































a 






























•o 






a> 

4> 






























c 
















4> 










> 








£ 


<U 




a 


^3 





T3 

C 
U 

s 

p 






f3 

-a 

"i 










a 


g 




id 
S 




13 


c 


-o 

1 




T3 

tn 

3 


*0 


1 


lA 




s 














(0 


S 





s 




5 

T3 


s 






Si 

en 


a 
c 


rt 


.2 


13 


c 











C 


_C 




.s 




.5 


s 


-a 
us 


a 


c 
tn 


LO 





G 




"73 








to 










LO 


rt 


3 ^ 






£ 


" 







c 


2 


c 




3 


a « 


3 


3 


O 








2 


3 


5 


3 c 





13 


3 


T3 

<u 


rt 


3 







COD -r 

.2 






rt 





rt 


3 


3 


3 


- 




T3 







T3 




^3 




13 XI 






Td 


^ 


•a 








X 




<u 




<u 


1) 














<u 




<u 














„ 


CC 




^ 


«tt 


w 


„ „ 


„ 


„ 




d 




„ 


„ 


„ 


^ vd 




'3 




<u 


'rt 




u 










'rt 





s *rt 








a 




fa 




P 


fa 




p 










fa 


Q 


fa 








P 


x 


CV3 


co 


On CO 


O 


r-^ 


r-. 


M 


v£) £5 


O 


t^. 


CO NO 


ro no 


NO 


_, 


r^ 


N CO 

ON NO 


to 


00 






ci 


O 


LO 


Cn 


O 




M 


w 


r^ n 




r^ 


8 


e 






ro 


On 


-<t 


O 


CO 


M 


-«*■ 


O 


On CO 


On CO 


ro 


On 


O WO 


1) 


N 


N 




N 


ro 




N 


M M 


M 




1-1 




>-i M 


N 




n 


N N 


CO 






































« 




O 


O 





§ 


O 





O 


8 8 



O ON 


O 


O 


O 


O 


O O 


O 


O 


8 

On 


O O 


i 


a 



<: 


it. 


ft 




8 


O 







O 


§ 


4 


I 


8 



10 O 


ON 


ro 


n On 




NO 


r^ 


Tt- 







n- 


r^ 


LT) \D 


Tt 


ro 


LO 


NO Tf 


fO 


O 


r^ 


On lo 


5 










N 
















M N 


C) 


N 


M 


N ro 


J 8 


S ° 


3 






s 


O 

u-i 


^ 


O 

M 


O O 

LO CO 


O 

O 


a 


O 

N 


% 


O O 





O 
O 


O 


NO CO 


1 4 


10 


to 


58 


CO 


10 


4 


N ON 


6 


vS" 


r^» 


r^ 


co *h 


CO 


ro 


ON 


ON OV 


CO 


Sf ^ 


r-^ 


t-. 


NO 


r^. 


r-» 


r^ vO 


r^ 


CO 


CO 


00 


O 


O 


ro 


ro ro 




■5 


CO 


LO 


C-> 


Cn 


r^ 


LO 


_, 


O NO 


rj- 


r^ 


CO 


vS 


O Tj- 


Tt 


^8 


ro NO CO 




. lo 


r^ 


nO 


N 


N 


r-^ 


MO 


W> CO 


ro 


ro 


LO 


NO NO 


NO 


4 


t*- ^J- 




U 


.a ck 


o\ 


o\ 


cK 


o\ 


ON 


o\ 


On On 


On 


ON 


10 


LO 


in io 


LO 


WO NO 


NO NO 


1 




Q 






































co 


LO 


ON 

LO 


CO 

en 


On 
co> 




r^ 


vq ^~ 


ON VO 

1" ^ 


^s 


vq 


vq no vq no 


tJ- 


%% 


S 


• a r>» 


r^. 


r^. 


rA 


r-^ 


r^> 


r^» 


tN. t^. 


r^» 


r^ 


LO 


LT] 


tn no 


NO 


NO 


co 


06 CO 




N 


,_, 


Tj" VO 


LO 


O 


O 


CO O 


On 


^. 


r> 


CO 


8 8 


3: 


8 


ro 


O ro 


co 


M 

c 


a *! 


N 


w 


N 


M 


N 





1- N 


CO 


N 


O 





On 


O 


O ON 




— CO 


CO 


d 


d 


d 


4- 


4 


4 co 


CO 


CO 


d 


d 


d 


z 


d 


d 


O w 




* ^ 


NO 











ro 


CO 


fO NO 


VO 


NO 


^o 


u-> 


10 ""> 


TT 


LO 


LO 


WO ■* 






1-1 


N 


M 


CI 


M 


D 


N M 


N 


^ 
















&U X 


tn C 


































Avers 

Rat 

of 

Grow 


w>-- CO 


1 


1 


1 


| 


1 


1 1 


1 


1 





LO 


m r-» 


O 


r^. 


NO 


WO wo 




























H 








x. 


LO 





q 


q 


q 


q 


q 


O O 


q 


O 





q 


q q 


q 


O 


q 


q q 


M 


£ O 


4 


CO 


10 


LO 


N 


1-^ 


O r^. 


r^ 


4 


d 


rA 


r-^ no 


LO 


\d 


r^. 


r^ r^. 




£ — 


r-^ 


f^ 


CO 


'ON 


un 


tr> 


tT ri 


O 


■x) 





NO 


NO lO 


On 


r^ 


LO 


95 


^ 


' N 


«s 


r) 


N 


N 


ro 


ro 


ro rf 


ro 


ro 


ro 


M 


N ro 


to 


CO NO 


Ov- 'S 


_ 


N 


ro 


■* 


un 


ON 


O 


HH NO 


r^ 


CO 


ri 


ro 


Tf On 


8 


h _ ( 


LO VO 1-^ 


£ O,<0 


OS 


C\ 


ro 


ro 


ro 


ro 


Tf 


■* ro 


ro 


ro 


O 


O 


O ON 

t-^ no 








O O 




f 


wo 


LO 


>-o 


LO 


LO 


LO 


"I 


W) 10 


LO 


u-1 


r^. 


t^ 


r^ 


t-^ 


r^ 


t^ t^ 



5H 



APPLIED MECHANICS. 



COMPRESSION OF WHITE PINE.— Single Sticks and Built Posts. 
In the multiple ones, dimensions of each stick are given. 









Dimensions of Post. 




Compression in 
150 In. Load 
= 500 lbs. per 
Sq. In. 


Ultimate Strength. 


No. 

of 

Test. 


Weight. 


*> ■£ a. 

£0.5 








Sectional 
Area. 






c 
u 

(-1 




a. 

D 

Q 


a 
"5j 


Ih 




lbs. 




in. 


in. 


in. 


sq. in. 


in. 


lbs. 




664 


153 


11 


i77-5o 


4.48 


11-65 


52.2 


o-°545 


IIOOOO 


2IO7 


665 


143 


10 


180.00 


4.48 


n.6 4 


52.1 


O.IOIO 


81500 


1504 


666 


163 


5 


179.97 


4-47 


11.63 


52.0 


0.0895 


70000 


1346 


667 


228 


13 


180.00 


5-4o 


11.30 


61.0 


0.0505 


160000 


2623 


668 


193 


5 


179-93 


5.61 


"•73 


65.8 


0.0622 


156300 


2375 


669 


253 


5 


180.00 


5-64 


11.76 


66.3 


0.0608 


152300 


2297 


638 


l$l« 


11 


180.00 
180.00 


4.50 
4-5° 


11.60 
11.59 


S2J--4 


1 0.0670 } 

1 0.0750 ) 


200000 


I916 


639 
640 




1? 
IS 


180.00 
180.00 
180.00 
180.00 


4-52 
4.49 

4-53 
4-52 


11.66 
11.62 
"•59 
n-59 


^2! "4-9 

£11-*- 


\ 0.0645 I 

1 0.0670 ) 
( 0.1060 ) 

\ 0.0955 1 


212000 
149000 


202I 
I419 


642 


!$!«* 


rs 


179.98 
179.98 


5-57 
5.58 


n. 61 
11. 61 


&!■•»■♦ 


5 0.0770 1 
J 0.0390 1 


215000 


l66l 


643 
644 


I3!*» 


1? 


179.92 
179.92 
179.96 
179.96 


5-65 
5-6i 
5.60 
5.60 


11. 61 
11.62 

nil 


glfl'3- 

§:?!■*« 


( 0.0440 ) 

1 0.0600 $ 

i 0.0596 

/ 0.0690 ) 


261000 
257800 


*995 
1980 


648 


< 206) 

1 203 i <°9 


Is 


180.00 
180.00 


5.60 
5-6i 


n.72 
n.72 


grtf-- 


\ 0.0590 ) 

1 0.0700 ^ 


268000 


2042 


649 
650 




! 9 
4 

1 12 

1 9 


180.00 
180.00 
180 00 
180.00 


5.60 
5-6i 

5-6i 
5.61 


11. 71 
n.74 

n-75 
11. 71 


SSI** 
If:? I '3.-6 


1 0.0600 ) 
/ 0.0705 i 

\ 0.0530 ) 
/ 0.0885 \ 


277000 
240000 


. 2107 

1824 


645 


lal* 


15 


179.97 
179.97 


5-59 
5.60 


"•59 
11.60 


S:Ji-»-' 


\ 0.0560 ) 
1 0.0540 \ 


263200 


2028 


646 
647 


is (3* 


IS 


180.00 
180.00 
180.00 
180.00 


5-59 
5.60 
5.62 
5.62 


11. 61 
11.62 
11.62 
11.62 


§1 }■"»*« 


\ 0.0493 I 

1 0.0620 \ 
1 0.0630 i 
1 0.0700 ) 


249000 
248000 


1915 
1899 


678 


l5l«* 


11 


179.94 
179.94 


5.58 
5-57 


11.47 
"•45 


'g:!K« 


1 0.0529 ) 
} 0.0642 ) 


245500 


1921 


679 
680 


1^(399 

ISI* 


JI2 

1 7 
11 


180.00 
180.00 
180.00 
180.00 


5.62 
5.62 
5.60 
5.61 


11.76 
11.72 
11.72 
"•73 




{ 0.0664 \ 
1 0.0705 ) 
\ 0.0650 > 

I 0.0495 i 


249000 
278000 


1886 
2116 


663 


i3i«* 


1*4 

/ 6 


180.00 
180.00 


5.60 
5.63 


"•75 
"•75 


S:!i-3... 


1 0.0621 1 
t 0.0657 1 


300000 


2273 


676 


15 1* 


1" 


179.94 
179.94 


5.60 
5.61 


11. 71 
"•73 


g:i I — 


f 0.0530 1 
t 0.0593 f 


274500 


2089 


677 


IS I* 


IS 


180.00 
180.00 


5.61 
5.68 


11.72 
11.72 


fell-- 


j 0.0551 J 

1 0.0625 j 


255000 


1945 



COMPRESSION OF WHITE PINE. 



515 



COMPRESSION OF WHITE PINE. — Concluded. 
Single Sticks and Built Posts. 







*o 
V "o 


Dimensions of Post. 




Compression in 
150 In. Load 
= 500 lbs. per 
Sq. In. 


Ultimate Strength. 


No. 

of 

Test. 


Weight. 


£0.s 








Sectional 
Area. 






.g 

M 

c 


•g 


■g 
ft 

<u 

P 


3 

< 






lbs. 




in. 


in. 


in. 


sq. in. 


in. 


lbs. 




690 


< 220 > 600 
(199) 


(18 


180.00 
180.00 
180.00 


4-52 
5.56 
4.46 


11.62 
11.70 
11.62 


52.5) 

65.0 \ 169.3 
51.8) 


I 0.0460 ) 
I 0.0580 > 
( 0.0480) 


310000 


1 831 


691 


(164) 
\ i97 { 520 
(159) 


1:1 


179.98 
179.98 
179.98 


4.48 
5.56 
4-45 


11.60 
11.60 
11. 61 


52.0 ) 

64.5 S 168.2 
51.7) 


( 0.0526 ) 
] 0.0430 [ 
( 0.0390 ) 


372500 


2215 


692 


lij 614 


S 


177-25 
I77-25 
I77-25 


5.62 

4-5° 
5.60 


11.60 
11.60 
".57 


65.2) 

52.2 > 182.2 

64.8) 


C 0.0580 ) 
< 0.0641 [ 
( 0.0768 ) 


363000 


1992 


687 


SI 536 


1" 

1 9 


180.00 
180.00 
180.00 


4-5o 
5.58 
4-52 


11.60 
11.62 
"•59 


52.2 1 

64.8 > 169.4 

52.4) 


( 0.0460 ) 

0.0587 
(0.0533) 


325500 


I919 


688 


\ 236 \ 57s 
(163) 


1" 

(11 


180.00 
180.00 
180.00 


4-5o 
4.62 
4-5o 


11.60 
11.60 
11.60 


52.2) 

65.2 > 169.6 

52.2) 


( 0.0565 ) 
j 0.0645 ) 
( 0.0703 ) 


306000 


1804 


689 


(188) 
j 203 \ 550 

( 159) 


K 


180.05 
180.05 
180.05 


4.48 
5.60 
4.46 


11.60 
11.62 
n-57 


52.0) 
65.1 1 168.7 


( 0.0510) 
< 0.0660 > 
( 0.0789 ) 


340000 


2015 


681 


fi93] 


L 9 


179-95 
179-95 
179-95 
*79-95 


4-47 
4-52 
4.48 
4- 5^ 


11.60 
n.65 

n.65 
11.65 


5i.9] 
52.5J 


f 0.0700) 
! 0.0714 ! 

1 0.0531 j 

1.0.0762 J 


362000 


1734 


682 


U51J 


{'5 

<- 9 


180.00 
180.00 
180.00 
180.00 


4-5o 
4-5o 
4.49 
4-5o 


n.63 
11.64 
11.60 
11.63 


52.3*1 

=52.4 i 

3 ^ > 209.1 

52.1 y 

S2.3J 


Co. 0612^ 
J 0.0546 ! 

] 0.0542 r 
(0.0916 j 


414000 


I980 


683 


f i69 i 

Si [787 

1 181 J 


fio 

11 


180.03 
180.03 
180.03 
180.03 


4.46 
5.62 
5.60 
4 46 


n.63 
11.69 
11.70 

11. 61 


5i. 9] 
5I-8J 


ro.o5 7 o , | 
J 0.0530 ! 

j 0.0494 r 
\ 0.0590 J 


501000 


2133 


684 


fx 4 61 

< 2IO Ws6 
1 234 j /3 
1 166 J 


{' 


180.00 
180.00 
180.00 
180.00 


4-50 
5-64 
5.60 
4.48 


11.64 
11-59 
n.58 
11. 61 


52.4] 

g:lf-M.« 

52. oj 


fo.06641 
J 0.0610 1 

) 0.0548 f 
V0.05I4J 


529000 


2255 


685 


fi45l 
! 220 1 

20 9 740 
I 166 J 




180.00 
180.00 
180.00 
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APPLIED MECHANICS. 



COMPRESSION OF YELLOW PINE. 
Single Sticks and Built Posts. 







"o . 


Dimensions of Post. 




Compression in 
150 In. Load 
= 500 lbs. per 
Sq. Inch. 


Ultimate Strength. 


No. 

of 

Test. 


Weight, 
in lbs. 


2 2 & 
£O.S 








Sectional 
Area. 










ft 
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3 


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in. 


in. 


in. 


sq. in. 








673 
673a 

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260 \ 


11 


( 180.05 
J 20.00 
( 20.00 


4.09 
4.01 
3-95 


n-35 
4.01 

3-97 


46.04 ) 
16.08 J 
15-68 J 


0.0370 


( 142200 
< 94000 
( 99600 


3065 
5846 
6352 


674 


233 


17 


180.00 


4-5i 


11.60 


52.30 


0.0492 


131500 


2515 


67S 


194 


22 


180.00 


4-34 


11.60 


50-30 


0.0490 


I2I200 


2410 


490 


269 


-• 


180.00 


5 -«>S 


12.10 


61.16 


- 


23OOOO 


3704 


670 


309 


12 


180.00 


5.65 


11.74 


66.30 


0.0455 


2059OO 


3106 


489 


287 


- 


180.08 


5-85 


12.05 


70.50 


- 


25OOOO 


3546 


654 


ISI«* 


i" 


180.00 
180.00 


5-63 
5.62 


11. 71 
11. 71 


^h- 8 


{ 0.0418 ) 
I 0.0540 i 


47OOOO 


3566 


655 


ISI*- 


ill 


179-93 
179-93 


S-64 
5.63 


11.72 
11.72 


&}■*■ 


( 0.0292 ) 
( 0.0315 i 


580OOO 


4387 


656 


l£l* 


111 


180.00 
180.00 


5.61 
5.61 


11. 71 
11. 71 


SSl™ 


I 0.0368 ) 
/ 0.0466 ) 


480OOO 


3653 


651 


l«i* 


i" 


180.00 
180.00 


5.58 
5.58 


11. 71 
11. 71 


Stl*** 


( 0.0620 ) 
( 0.0514 ) 


360OOO 


2756 


652 


ISil* 


{S 


180.00 
180.00 


5.58 
5.58 


11.70 
11. 71 


S3l"3*» 


\ 0.0395 ; 
j 0.0360 S 


5885OO 


45o6 


653 


l£l«« 


{2 


180.00 
180.00 


5.63 
5-59 


11. 71 
11.68 


SSI**- 


\ 0.0559) 
I 0.0550 J 


43660O 


3328 


657 


iiS!*» 


!.! 


179.96 
179.96 


5.63 
5-64 


11.72 
11. 71 


821 -3- 


{0.0375 { 
1 0.0305 ) 


580OOO 


4394 


658 


IgJ* 


1" 


179.98 
179.98 


5-59 
5-59 


11. 71 
11.72 


%.\W-° 


^ 0.0320 J 
) 0.0436 ) 


4480OO 


3420 


659 


ISl«* 


1" 


180.00 
180.00 


5-6i 
5-6i 


"•73 
"•73 


§:!!-*■« 


| 0.0312 ) 
t 0.0372 \ 


60OOOO 


4559 


660 


tSI»" 


is 


180.03 
180.03 


5-6i 
5.63 


11.22 
11.24 


SSI 1 -*' 


\ 0.0325 ; 
/ 0.0400 \ 


51OOOO 


4041 


661 


is I* 


I" 


180.00 
180.00 


5-66 
5.60 


11.70 
11.72 


66 - 2 U,t 8 
65.6 j?3*.8 


{ 0.0410 ) 
} 0.0365 \ 


4IOOOO 


3111 


662 


|S! 57- 


ft 


180.00 
180.00 


5-6i 
5.61 


"•75 
"•75 


65-9 s a 


l 0.0540 | 
I 0.0500 j 


3880OO 


2944 


693 


lit" 4 


(23 


179-97 
179.97 
179.97 


4-5° 
5-5o 
4.49 


11. 61 
11.56 
11.62 


52.2) 

63.6 1 168.0 
52.2) 


f 0.0320 ) 
{ 0325 J 
( 0.0148 ) 


564OOO 


3357 


694 


l* 93 h 

< 290 } 691 

(208) 


p 

(10 


180.00 
180.00 
180.00 


4-5o 
5-59 
4.46 


"•35 
11.36 

"•35 


51. 1) 

63.5 > 165.2 

50.6) 


( 0.0500 ) 
1 0.0650 1 
( 0.0610 ) 


500000 


3027 


695 


(207 J 

J 290 \ 743 
(246) 


{J 

(16 


180.00 
180.00 
180.00 


4.49 
5.20 
4-5° 


"•35 
"•34 
"•35 


51,0 I A 
59.0 > IOI.I 

5i-i) 


( 0.0429 1 
1 0.0290 > 
( 0.0410) 


474OOO 


2942 



COMPRESSION OF YELLOW PINE. 



521 



COMPRESSION OF YELLOW PINE.— Concluded. 
Single Sticks and Built Posts. 









t> 


Dimensions of Post. 




ft. 
-J . 


Ultimate Strength. 


No. 
of 

Test. 


Weight, 
in lbs. 


*4* 

£O.S 








Sectional 
Area. 


Compressior 
150 In. 
= 500 lbs 
Sq. Inch. 






£ 
ff 

^ 




0. 

V 

P 


3 
3 


U 

JDUl 

►J 








in. 


in. 


in. 


sq. in. 








696 


(217) 

j 253 J 660 


IS 

(15 


180.00 
180.00 
180.00 


4-5i 
5-5° 
4.48 


11.24 
11.23 
11.23 


50.7) 
61.8 [162.8 


( 0.0337 ) 
) 0.0567 \ 
(0.0715) 


480000 


2948 


697 


(242) 
j 249 j 706 


I s 

(15 


180.00 
180.00 
180.00 


4.52 
5-43 
4-47 


11.60 
11.60 
11.58 


52-4) 

63.0 > 167.2 
51-8) 


( 0.0240 } 
] 0.0330 J 
( 0.0336 ) 


540000 


323O 


698 


(224) 
\ 255 \ 775 
(296) 


r 1 


179.94 
179.94 
179.94 


4.46 
5-53 
4-5o 


11.60 
11.70 

11.59 


5i.7) 

64.7 | 168.6 
52.2 ) 


(0.0290) 
J 0.0385 [ 
( 0.0341 ) 


544000 


3227 


488 


911 


- 


180.20 


(6.88 
J6.72 


15-75 
15-75 


Io8 - 36 1 214 2 
105.84 j 2I4 ' 2 


- 


700000 


3268 



These sets of tests furnish us practically the only reliable 
information we have in regard to the strength of full-size col- 
umns of white pine, yellow pine, and oak. In regard to white 
and yellow pine, Mr. Edward F. Ely, instructor in architecture 
at the Massachusetts Institute of Technology, has plotted the 
average, and also the lowest, results of the tables ; and, from an 
inspection of the diagrams, he gives the following rules for 
determining the breaking-strength of a column with flat ends, 
the load being evenly distributed over the ends : — 

Let A = area of section in square inches. 

/= constant whose value is given in the tables fol- 
lowing. 

- = ratio of length to least side of rectangle, all the 

tests having been made on rectangular sections. 



Then 



Breaking-strength =/A, 



522 



APPLIED MECHANICS. 



where f has the following values : — 



White Pine. 


Yellow 


Pine. 


/ 
r 


/ 


/ 
r 


/ 


o to IO 


2500 


to 15 


4000 


10 to 35 


2000 


15 to 30 


3500 


35 t0 45 


1500 


30 to 40 


3000 


45 to 6o 


1000 


40 to 45 


2500 






45 t0 5° 


2000 






50 to 60 


1500 



In the case of oak, if it is desired to apply the results to 
greater ratios of length to diameter than those tested, a similar 
reduction can be made in the value of f to that which takes 
place here in the case of white and yellow pine. 

For the ratios of length to diameter tested, f = 3000 would 
seem to be a fair value to use. 

§ 238. Factor of Safety. — Whereas we are constantly told, 
that, in the case of iron bridge-work, we should use a factor of 
safety 4, but that for a timber construction we should use at 
least 8, or even 10 and 12, — the reason evidently is, that the 
figures that have generally been given us for breaking-strength 
have been really from two to four times the actual breaking- 
strength, — it would seem to the writer, that, when we use 
correct values for breaking-strength, a factor of safety 4 will 
be sufficient for all ordinary timber constructions ; i.e., that we 
should use for working-strength per square inch, one-fourth 
the breaking-strength per square inch. This same reasoning 
will also apply to the case of beams bearing a transverse 
load when they are designed with reference to their breaking- 
weight. 



TRANSVERSE STRENGTH OF TIMBER. 



523 



§ 239. Transverse Strength of Timber. — In this regard, 
the common theory of beams has already been explained in 
§ 1%$ et seq. 

The tables of Rankine and Rodman, already given, represent 
the values of modulus of rupture that have been in common 
use. Other values, not differing essentially from these, are given 
by Hatfield, Laslett, Thurston, Trautwine, and others, all based 
upon tests of small pieces. No tables of these values will be 
given ; as those above referred to furnish practically the same 
information. Confining ourselves to tests of full-size pieces, we 
find an account of a set of tests attributed by D. K. Clark, in 
his " Rules and Tables," to Edwin Clark and C. Graham Smith. 
The results are given below, and it will be seen that they are 
very much below those given by experimenters on small pieces. 
Two tests by R. Baker are also mentioned by D. K. Clark. 



Kind of Timber. 


Breadth 

and 
Depth. 


Span. 


How 
Loaded. 


Breaking- 
Weight. 


Modulus 

of 
Rupture. 




in. 


ft. 








American red pine 


I2.0 X I2.0 


15.OO 


Centre 


33497 


5238 


<< << « 


I2.0 X 12.O 


15.OO 


M 


29908 


4680 


« ti « 


6.0 X 6.0 


7-50 


it 


737o 


4608 


Memel fir . . . 


13-5 X 13-5 


10.50 


Distributed 


68560 


5 2 74 


<< « 






13-5 X 13-5 


10.50 


<< 


68560 


5274 


Baltic fir . 






6.0 X 12.0 


12.25 


Centre 


*9HS 


4878 


« << 






6.0 X 12.0 


12.25 


11 


23625 


6020 


Pitch pine 






6.0 X 12.0 


12.25 


U 


23030 


5868 


« K 






6.0 x 12.0 


12.25 


" 


23700 


6048 


«i <( 






14.0 X 15.0 


10.50 


<< 


134400 


8064 


a u 






14.0 X 15.0 


10.50 


(t 


132610 


795 6 


Red pine . 






6.0 X 12.0 


12.25 


" 


16800 


4284 


it h ( 






6.0 X 12.0 


12.25 


" 


19040 


4860 


Quebec yellow pine 


14.0 X 15.0 


10.50 


Distributed 


68600 


4122 


U it M 


14.0 X 15.0 


10.50 


" 


68600 


4122 


tt tt it 


14.0 X 15.0 


10.50 


Centre 


85792 


5148 


it it it 


14.0 X 15.0 


10.50 


«< 


76160 


4572 



524 APPLIED MECHANICS. 

During the last three years tests of the strength and stiff- 
ness of full-size beams of spruce, yellow pine, oak, and white 
pine, both under centre loads and distributed loads, have been 
carried on in the Laboratory of Applied Mechanics of the 
Massachusetts Institute of Technology. Tests have also been 
made upon the effect of time on the stiffness of such beams, 
also on the strength of built-up beams, and of floors and fram- 
ing-joints, all full size. A summary of the results obtained 
will be given, and conclusions drawn as to the proper values 
of the modulus of rupture and modulus of elasticity, etc., to 
be used in practice. 

Before giving this summary, the following explanation and 
formulae will be appended for the convenience of those who 
may not have read the former part of this book. The different 
tables of results given in different handbooks differ in the form 
of the constant. Thus, the constant given by Trautwine and 
Hatfield is one-eighteenth the modulus of rupture, or the hypo- 
thetical breaking centre load of a beam one inch square and 
one foot long supported at the ends ; while Rodman gives one- 
sixth of the modulus of rupture, or the hypothetical breaking- 
load at the end of a cantilever one inch square and one inch 
long. 

The formulas for breaking-load in terms of the modulus of 
rupture, and for modulus of rupture in terms of the breaking- 
load, for some of the most usually occurring cases of rectan- 
gular beams, are appended. 

Let W = breaking-load in pounds. 

f — modulus of rupture in pounds per square inch. 
b = breadth of beam in inches. 
h = depth of beam in inches. 
/ = length of beam in inches. 

Then we shall have : — 
(a) Beam fixed at one end and free at the other. 



TRANSVERSE STRENGTH OF TIMBER. 525 



i°. Single load at free end, 

7 6/' J ~l^ 



2°. Load uniformly distributed, 



J 3/' y M^ 



(£) Beam supported at both ends. 
i°. Single load at the middle, 



2°. Load uniformly distributed, 

* 1' J 4M 2 

3 . Single load at a distance a from the origin, 

__^_ 6Wa(/-a) 

J 6a{l- a)' J Ibh* 



DEFLECTION OF BEAMS. 

While the preceding formulas refer to the breaking-strength 
of beams, it is better engineering to determine, as the safe load 
of a timber beam, the load that will not deflect it more than a 
certain small fraction (3^ or -^\^) of the span. 
Let W = given load in pounds. 
b = breadth in inches. 
h = depth in inches. 
/ = length in inches. 
v == greatest deflection in inches. 
E = modulus of elasticity of the material in pounds 
per square inch. 



526 APPLIED MECHANICS. 

(a) Beam fixed at one end and free at the other. 
i°. Single load at free end, 

4 073 4 jv/3 

1) = , Jz, = . 

Ebfc vbh* 

2°. Load uniformly distributed, 

^3 W*\ E = *^. 

2 Ebk? 2 z/M 3 ' 

(£) Beam supported at both ends. 
i°. Single load at the middle, 

_ i Wfi_ _ i 07_ 3 

4 isM 3 ' 4 vbfa 

2°. Load uniformly distributed, 

= J_^3_ = 5 073 

32 Eb/i 3 ' 32 #M 3 

The above formulae enable us to determine the deflection 
of a beam under a given load when the modulus of elasticity 
of the material is known, or to determine the modulus of elas- 
ticity of the material from the observed deflection. 

LONGITUDINAL SHEARING. 

In any rectangular beam, the greatest intensity of the lon- 
gitudinal shearing-force at any section (i.e., its intensity at the 
neutral axis of the section) is, if we let 

F = shearing-force at the section, technically so-called, 

in pounds, 
b = breadth of beam in inches, 
h = depth of beam in inches, 

ZF 
2ih' 



TRANSVERSE STRENGTH OF TIMBER. 



527 



In the case of a beam supported at the ends, and loaded at 

the middle with a single load, we have, for all sections except 

W 
the middle, F = — ; and hence, if we denote by 1 the greatest 

intensity of the longitudinal shearing-force at the neutral layer, 
we shall have 

3 if. 

4 bh 






<o 



and this is the intensity of the shearing-force at all points 
along the neutral layer, except at the middle section, where it 
is zero. 

In the case of a beam supported at the ends, with the load 
uniformly distributed, the greatest intensity is that at the sup- 
port, and is also given by equation (1), but decreases gradually 
to the middle. 



SUMMARY OF THE TESTS. 

The tests recorded may be divided into six classes : — 
i°. Spruce beams. 4 . Oak beams. 



2 . Yellow-pine beams. 
3 . Time tests. 



5 . White-pine beams. 
6°. Framing-joints. 



1°. Spruce Beams. — Before giving a summary of the tests 
made in this laboratory, I will insert some of the moduli of 
rupture and moduli of elasticity given by different authorities. 

Moduli of rupture are given as follows : — 





Maximum. 


Minimum. 


Mean. 


Hatfield .... 


I2996 


7506 


99OO 


Rankine .... 


12300 


9900 


1 1 IOO 


Laslett 


9707 


7506 


9045 


Trautwine .... 


- 


- 


8lOO 


Rodman .... 


- 


- 


6168 



528 APPLIED MECHANICS. 

Hatfield's, Laslett's, Trautwine's, and Rodman's figures are 
from their own experiments. Trautwine advises, for practical 
use, to deduct one-third on account of knots and defects, hence 
to use 5400. The tables show the values obtained in these 
tests, and I will add a recommendation as to the values of 
modulus of rupture and modulus of elasticity suitable to use in 
practice. 

As a result of the tests thus far made in my laboratory, 
it seems to me safe to say, if our Boston lumber-yards are 
to be taken as a fair sample of the lumber-yards in the case 
of spruce, — if such lumber is ordered from a dealer of good 
repute, no selection being made except to discard that which is 
rotten or has holes in it, — that 3000 lbs. per square inch is all 
that could with any safety be used for a modulus of rupture, 
and even this might err in some cases in being too large ; 
(2 ) that, if the lumber is carefully selected at any one lumber- 
yard, so as to take only the best of their stock, it would not be 
safe to use for modulus of rupture a number greater than 4000 ; 
and if we required a lot of spruce which should have a modulus 
of rupture of 5000, it would be necessary to select a very few 
pieces from each lumber-yard in the city. With a factor of 
safety four, we should have for greatest allowable outside fibre 
stress in the three cases respectively, 750, 1000, and 1250. 

The modulus of elasticity (i.e., that determined from the 
immediate deflections) was: maximum, 1588548; minimum, 
897961 ; mean, 1329479. 

Two time tests were made on yellow pine ; and, if we should 
consider the effect of time on spruce the same as for yellow 
pine, we should obtain for use, for spruce, about 886319. 



TRANSVERSE STRENGTH OF TIMBER. 



529 



















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APPLIED MECHANICS. 






8 « * 



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TRANSVERSE STRENGTH OF TIMBER. 



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532 



APPLIED MECHANICS. 



Yellow-Pine Beams. — The moduli of rupture in common 
use are given as follows by different authorities ; viz., — 





Maximum. 


Minimum. 


Mean. 


Hatfield .... 


2II68 


9OOO 


15300 


Laslett 


1 41 62 


IOO44 


I2254 


Trautwine .... 


- 


( Yellow pine 
( Pitch pine 


9000 
990O 


Rodman .... 


9876 


8796 


9293 



A summary of the figures obtained from these tests will be 
given in a table at the end of these remarks. 
It will be observed that we have for 





Maximum. 


Minimum. 


Mean. 


Modulus of rupture . . 
Modulus of elasticity . 


9380 

2386096 


4764 
I256286 


6984 
1779517 



We also have a considerable reduction of the modulus of 
elasticity with time, as shown by the time tests ; the proper 
values for use being, in the case of spruce and hard pine, from 
two-thirds to three-fourths the immediate moduli of elasticity. 
(See p. 536.) 



TRANSVERSE STRENGTH OF TIMBER. 



533 



YELLOW-PINE BEAMS. 



No. 

of 

Test. 


Width and 
Depth. 


Span. 


Manner of Loading. 


Breaking- 
Weight, 
in lbs. 


Modulus of 
Rupture. 


Modulus of 
Elasticity. 




inches. 


ft. in. 






. 




30 


3 x i3i 


14 


Load at centre 


I S I 5 8 


6614 


1937025 


32 


4TTT X 12-^6 


18 O 


« << 


I375I 


7383 


1733976 


33 


3ii x «i 


I'8 O 


U it 


9832 


5386 


1793923 


47 


3 x i 3 | 


14 


it a 


19574 


8696 


2386096 


50 


4 X i 4l V 


21 O 


a a 


12875 


59H 


1256286 


53 


3t x 14 


24 6 


a a 


10076 


7206 


I784426 


54 


3 x I2i 


24 


it tt 


9576 


9380 


2II682I 


56 


3i x 14 


15 4 


u a 


10572 


4764 


1490396 


57 


2\% X 12 


19 2 


tt tt 


8472 


6950 


I444521 


59 


9 x i 3 3 


24 


u it 


21083 


5352 


I4I7793 


62 


4i x i2& 


19 10 


it a 


1 546 1 


9102 


2037939 


63 


4i% X 12A 


20 


" " 


14073 


8145 


1599339 


64 


4i X 12} 


19 10 


it tt 


!°573 


6098 


1917976 


65 


4 X I2i 


19 8 


it a 


II573 


6782 


I9667 I 7 


67 


4 i X 12 


18 6 


a a 


13374 


7277 


' 1787610 


68 


4 X I2i 


19 9 


tt it 


17676 


10872 


2381685 


69 


3A x 14 


20 


tt tt 


6675 


3963 


1 169298 


7i 


4i X 12 


18 2 


tt tt 


16074 


8248 


1512192 


74 


4 X 12 


20 


it a 


1 107 1 


7004 


1628 I 34 


75 


4 X iif 


19 9 


tt tt 


i377i 


939 1 


1850667 


76 


4i X 12-^ 


17 4 


( Load equally ) 
1 distributed > 
( at 1 2 points ) 


15825 


4207 


1344083 


77 


4-} X 12 


17 4 


( Load equally ) 
) distributed > 
( at 12 points ) 


373 2 5 


10286 


2 I 23I 54 


78 


4 X I2| 


22 10 


Load at centre 


7172 


4845 


I455308 


79 


4 X 12 


19 8 


" " 


- 


- 


2087583 


81 


4? X I2i 


17 4 


( Load equally ) 
1 distributed > 
( at 1 2 points) 


16025 


4349 


I I 62467 


82 


4 X 12 


19 8 


Load at centre 


I5S7I 


9671 


1607336 


84 


4i X I2i 


21 4 


a 


1 1374 


6985 


I 501854 


85 


4 X Il| 


20 6 


it a 


16874 


1 1360 


2246154 



534 



APPLIED MECHANICS. 



YELLOW-PINE BEAMS.— Concluded. 



No. 

of 

Test. 


Width and 
Depth. 


Span. 


Manner of Loading. 


Breaking- 
Weight, 
in lbs. 


Modulus of 
Rupture. 


Modulus of 
Elasticity. 


87 

88 

9i 

92 


inches. 

4 X ii\ 

6 X li\ 

4 ,X 12 

6 x 12 

Average 
Average 


ft. in. 
21 4 
20 4 
19 IO 

6 5 
modulus 
modulus 


Load at centre 

u it 

u « 

U it 

of rupture 


II272 

J 5283 
18074 
38090 


7335 
61 12 

1 1303 
5092 


1535647 
1613OI2 
2223795 


31)226072 


7292 


of elasticity . 




31)54113213 


I7455 8 7 









In regard to the modulus of rupture to be used in practice 
for yellow pine, I should say, that, for the modulus of rupture 
of yellow pine of fair quality, I should not feel justified in using 
a number greater than 5000 lbs. per square inch, especially for 
large sizes, such as 9 X 14 inches, 12 X 16 inches, etc. My 
reason for this conclusion is, that, although the average modulus 
of rupture derived from the tests already enumerated is 7292, 
nevertheless, we have, in the case of beam No. 59, a modulus 
of rupture of 5300, notwithstanding the fact that this beam 
was quite free from knots, cracks, crooked grain, and other 
defects, and had been selected by a builder as one of excep- 
tionally good quality. With a factor of safety four, we should 
have about 1200 as our greatest allowable outside fibre stress. 

3 . Time Tests. — Two time tests have been made on 
yellow-pine beams. 

In the first the beam was 4 X 12.5 inches, 20 feet span: 
it remained under load from Nov. 15 to Jan. 8. It was loaded 
at the centre, at first with 485 lbs., which was increased on 



TRANSVERSE STRENGTH OF TIMBER. 



535 



Nov. 21 to 1289, on Nov. 25 to 2093, etc. ; finally breaking with 
11741 lbs., giving a modulus of rupture of 6742 lbs. per square 
inch. The modulus of elasticity determined from the respec- 
tive deflections was as follows : — 



From immediate deflection, 172160S 
At 4103 lbs 1377166 



6113 
7118 
8123 



1293046 

-"5793 1 
1169451 



At 8525 lbs 1 1 13684 

" 9329 " 1101809 

" 10133 " 1108016 

" 10334 " 1 103987 



In the second time test, the beam was 4X12 inches, 21 feet 
6 inches span, loaded at the centre. 

It was loaded on* Dec. 28 with 2070 lbs., which was increased 
to 3070 on Jan. 1 ; this load remaining on the beam till Jan. 30, 
when the load was increased until the beam broke at 11770 lbs., 
giving a modulus of rupture of 8019 lbs. per square inch. 

The modulus of elasticity determined from the immediate 
deflection was 171 5880 lbs. per square inch; that determined 
from the final deflection at 3070 lbs. was 13439 18 lbs. P er square 
inch, or about two-thirds the immediate modulus of elasticity. 

A third time test, made with a distributed load on a spruce 
beam, has shown a similar result. 



PROPER VALUE OF MODULUS OF ELASTICITY FOR USE IN 
COMPUTING DEFLECTIONS. 

The fact that the strength of a structure is the strength of 
its weakest part, should lead us to select for use in ordinary 
cases, for immediate modulus of elasticity, a number less than 
the average. 

Whether it is best, in any particular case, to use a number 
any greater than the minimum, I leave the reader to determine 
from the circumstances of the case and a perusal of the tests. 
Moreover, it should be distinctly understood, that, when the 
immediate modulus of elasticity is used in computing deflec- 
tions, the deflections are those that will be assumed by the 






536 



A P PLIED ME CHA NICS. 



beam immediately after the application of the load ; while the 
deflections of the beam after the load has remained on it for a 
certain length of time will be' greater. 



MODULUS OF ELASTICITY TO BE USED IN COMPUTING THE DE- 
FLECTIONS OF BEAMS AFTER THE LOAD HAS BEEN ON FOR 
SOME TIME. 

A perusal of the time tests tends to show, that, in com- 
puting the final deflection of a beam under a given load, we 
ought to use a modulus of elasticity no greater than three- 
fourths (and two-thirds would be safer) of the immediate 
modulus of elasticity ; and it will be noticed that the values 
of the immediate moduli of elasticity deduced from these tests 
are a little, though not very much, smaller than those given by 
Rankine, Trautwine, Hatfield, etc. 

4°. Oak Beams. 

SUMMARY. 



No. 

of 

Test. 


Span. 


Width and 
Depth. 


Description. 


Breaking- 
Weight, 
. in lbs. 


Modulus of 
Rupture. 


Modulus of 
Elasticity. 


48 

5 1 
55 
80 


ft. in. 
19 6 
15 6 
13 8 
18 

Aver 

Aver 


inches. 

6 x 12 

4i x Hi 

3 x 13! 

4 X 12 
age modulus 
age modulus 


Load at middle 

<< (< K 
II II (( 
II II II 


13776 
I9076 
1 067 1 

x 337i 


5596 
6060 
4984 
7659 

4)24299 


1766839 

I240728 

853098 

1 307 1 80 


6075 






4)5169836 


1292459 









While the average modulus of rupture is 6075, this is evi- 
dently too high a value to use in practice. I leave the reader 
to judge, but I should not feel safe with more than 4000 lbs. 
per square inch. 



TRANSVERSE STRENGTH OF TIMBER. 



537 



«J°. White-Pine Beams. 









Breaking 








No. of 


Width and 


Span. 


Centre 


Modulus of 


Modulus of 




Test. 


Depth. 


Load, in 


Rupture. 


Elasticity. 










lbs. 










inches. 


ft. in. 










• 94 


3 x "i 


15 8 


5088 


3613 


924252 


( Pattern stock. 


95 


3 x 13 


14 


I2588 


7251 


I280832 


< Clear piece. 
1 Seasoned 3 yrs. 


96 


3 x 13 


16 6 


9088 


53 2 4 


IO72889 




97 


3 X II 


15 8 


6088 


4729 


978256 




98 


2| x 9 f 


16 


6088 


64IS 


1234880 




99 


2| X 13 


15 6 


59 88 


3438 


IO20390 




100 


3 x 9! 


16 


4288 


4330 


I 165937 




102 


3 x lof 


15 6 


4790 


3355 


990190 




103 


3 x 11 


16 6 


6588 


5390 


1242649 




104 


3 x ni 

Average 


15 6 

modulus 


5088 
rupture . 


3739 


930760 




48084 


10842035 


. 4808 






Average 


modulus 


elasticity 




I0842OO 





These averages become respectively, if 95 be omitted, 
1062356 and 4537. 

It would seem to the writer, therefore, that the rule already- 
laid down in the case of spruce would apply also to white pine. 



LONGITUDINAL SHEARING. 

Below are given tables showing the greatest intensity of the 
shear at the neutral axis of each beam at the time of fracture. 
I will give tables showing these results : and we must observe, 
that, in the case of those beams which gave way by shearing, 
the figures given represent the shearing-strength of the wood 
along the grain ; while in the case of those that did not give 
way by shearing, it is fair to assume that these numbers are 
less than the shearing-strength of the wood. 



533 



APPLIED MECHANICS. 



TABLE OF BEAMS THAT GAVE WAY BY LONGITUDINAL SHEARING. 



Spruce. 


Yellow Pine. 


No. 


Width 

and 
Depth. 


Span. 


Intensity 

of 

Shear. 


No. 


Width 

and 
Depth. 


Span. 


Intensity 

of 

Shear. 


22 
24 
31 

35 
36 
46 


3 f x 12 
3 x 12 
3i x 12 
6 x 12 

2 X ll| 
3! X 12 

Average . 


ft. in. 
14 O 
14 
18 O 
18 
7 2 
10 2 


202 
I90 

J 54 
117 
248 
233 


30 
32 

33 
50 
92 


3 x 131 
4fV X i4ft 

3t£ x I2 4 

4 X i 4l -ig 
6 x 12 

Average . 


ft. in. 
14 O 
18 
18 

21 O 

6 8 


273 
242 

*53 

172 

397 


6)1144 
191 


5) I2 37 
24S 



TABLE OF BEAMS WHICH DID NOT FAIL BY SHEARING. 



Spruce. 


Yellow Pine. 


No. of Test. 


Maximum Intensity of 
Shear at Fracture. 


No. of Test. 


Maximum Intensity of 
Shear at Fracture. 


3 


l8l 


47 


359 


4 


3 OI 


53 


179 


5 


174 


54 


203 


6 


230 


56 


185 


7 


308 


57 


182 


8 


170 


59 


133 


9 


141 


62 


231 


10 


106 


^3 


211 


11 


208 


64 


161 


12 


126 


65 


183 


14 


I05 


67 


196 


15 


304 


68 


273 


16 


277 


69 


112 


17 


465 


7i 


223 


18 


202 


74 


173 



TRANSVERSE STRENGTH OF TIMBER. 



539 



TABLE OF BEAMS WHICH DID NOT FAIL BY SHEARING.- Concluded. 



Spruce. 


Yellow Pine. 


No. of Test. 


Maximum Intensity of 


No. of Test. 


Maximum Intensity of 




Shear at Fracture. 




Shear at Fracture. 


19 


138 


75 


230 


20 


160 


76 


231 


21 


137 


77 


549 


23 


I08 


78 


108 


25 


123 


81 


238 


26 


155 


82 


343 


. 27 


167 


84 


164 


28 


134 


85 


270 


29 


129 


87 


172 


37 


169 


88 


156 


45 


133 


9i 


282 


49 


205 






60 


406 






66 


228 






70 


206 






72 


174 






90 
Average . . 


272 


Average . . 




199 


221 



One would naturally expect to find the intensity of the 
shearing-stress at fracture less in the case of the beams that 
did not fail by shearing than in the case of those that did ; and 
this is seen to be generally true (making allowance for different 
qualities) both in the case of spruce and hard pine. 

The notable exceptions seem to be, in the case of spruce, 
beams Nos. 4, 7, 15, 16, 17, 60, 90, all of which have this 
intensity very large. If these be omitted from the list, the 
average for those that did not give way by shearing would be 
160 pounds per square inch, which is less than 191, the average 
for those that did. 



540 APPLIED MECHANICS. 

In the case of yellow pine, the notable exceptions are beams 
Nos. 47, JJ, and 92 ; and, if these be omitted, the average for 
those yellow-pine beams that did not fail by shearing would be 
197 pounds, which is less than 248. 

Moreover, it is to be observed, that, in the case of the spruce, 
Nos. 4, 7, 15, 16, and 17 were all of smaller dimensions than 
those used in practice. 

In the face of these apparent exceptions, which I am unable 
to explain, I prefer not to state at present any definite rule for 
the guidance of one who wishes to take the shearing-force into 
account in his calculations, but rather to leave him to use his 
judgment, in connection with these results, for any particular 
case. 

It will also be observed that these shearing-forces are less 
than those obtained from the experiments on direct shearing 
along the grain, made at the Watertown Arsenal ; and this is 
naturally to be expected, for the shearing in their case took 
place along a section that was perfectly sound, while in these 
cases it took place at the weakest point. 

6°. Framing-yoints . — Another matter intimately connected 
with the strength of timber beams is the strength of the beam 
after it has been cut in some of the various ways commonly 
employed in framing. We are often told that a notch cut on 
top of a beam, or at the middle of its depth, or near the sup- 
port, does but little injury ; but the tests made, show the injury 
to be very large, amounting to a reduction of the strength of 
the beam to one-fourth or one-fifth of its original strength, 
with some of the most approved framing. The fact is, that, 
with a material where the shearing-strength along the grain is 
so small as it is in the case of timber, almost any cutting does 
a great deal of injury; and it is much better to avoid framing 
whenever it is possible, and use stirrup irons instead. In these 
tests, only two of the most approved framing-joints have been 
tested; viz., the joint known as the " tusk-and-tenon," shown 



TRANSVERSE STRENGTH OF TIMBER. 



541 



in Fig. 243, and used for framing the tail-beams of a floor into 
the headers, and the " double 
tenon and joint bolt," shown in 
Fig. 244, and used for framing 
the headers into the trim- 
mers. 

The arrangement is shown in plan in Fig. 245, where I and 
2 are the trimmers, 3 is the header, and 4, 5, and 6 are the 
tail-beams ; the latter being supported at 





Fig. 243. 



Fig. 244. 



^ >" 


r ( 


i 


r- 


> 4 









\ 5 


10 




3 


S 6 


"3 




m 


/ 


!- ( 



one end on the header, and at the other 
on the wall, the header being supported 
by the trimmers, and the trimmers being 
supported on the walls at both ends. 

It is sometimes the practice to hang 
the header in stirrup irons, and this is an 
improvement; but it is very seldom that 
the tail-beams are hung in stirrup irons, 
and these tests have shown the weakening 
already referred to, from the mortises cut 
in the header to admit the tail-beams. 

In our earlier tests, these joints were 

tested by loading, the header at the top, 

FlG * 245- distributing the load over the mortises ; 

and the headers then tested were made for four tail-beams. The 

results are given below, these headers being all of spruce : — 



542 



APPLIED MECHANICS. 



No. 




Width 




Total 


of 


Span. 


and 


Description. 


Breaking- 


Test. 




Depth. 




Weights. 




ft. in. 


in. 


f Header. Framed at ends. Mor-' 




2 


6 8 


4 X 12 


•j tised for four tail-beams. Loads • 
[ applied above mortises. 


IO338 
















f Header. Framed at ends. Mor-") 




38 


6 8 


6 X 12 


\ tised for four tail-beams. Loads \ 
[ applied above mortises. J 
f Header. Hung in stirrup irons. 1 


IO798 


39 


6 8 


6 X 12 


•j Mortised for four tail-beams. \ 
{ Loads applied above mortises. J 
( Header. Framed at ends. Mor- ] 


21298 


4* 


6 8 


3 i x 12 


•j tised for four tail-beams. Loads f- 
[ applied above mortises. J 


10757 











It is evident that the breaking-weights found here are 
greater than those that would actually break the header when 
used in a floor ; for in the latter case the load is not applied at 
the top of the header, but lower down, and brings about an 
additional tendency to split the header. 

A spruce floor was next built and tested, the following being 
a partial account of the test : — 

No. $2. — Section of a floor between the trimmers. Spruce : 
three tail-beams, 2 inches by 12 inches each, framed into a 31- 
inch by n|-inch header; header in turn framed into sections 
of the trimmers by double tenon and joint-bolt, cross-bridged in 
two places; tail-beams framed by tusk-and-tenon joint, pinned, 
floored over and furred below ; load at centre, distributed be- 
tween the three tail-beams by bridging. 

Span = 16 feet ; weight of joist, flooring, etc., = 331 lbs. 

1 1238 lbs. = breaking-load. 

Joist on east side broke by splitting off at the tenon, bore 
7988 lbs. after. The load was then increased. Centre tail-beam 



TRANSVERSE STRENGTH OF TIMBER. 543 

broke by tension at 9988 lbs., on account of cross-grain in the 
lower fibres. A split also started at the lower tenon of the 
header," which at the time of breaking was rapidly increasing. 

Average modulus of rupture of the tail-beams, including 
their own weight, etc., == 3801 lbs. per square inch. 

Average modulus of elasticity of tail-beams == 1399141 lbs. 
per square inch. 

It is to be noticed, that the header already began to crack 
when the tail-beams broke, and hence that the floor could have 
borne but little more, even if the load had been uniformly dis- 
tributed : hence that, in this case, the breaking-strength of the 
floor would be determined by calculating the loads at the centre 
of the tail-beams, instead of accounting it as distributed ; in 
other words, the breaking-weight would be about one-half what 
we should get by considering the load as distributed on the 
tail-beams. Since that time we have had six tests on yellow- 
pine headers, which will be given here. 

It will be seen from these tests, that the first of these head- 
ers had for its breaking-weight 10916 lbs., and the second 
1 3 163, or in each case one-half the load on the floor. To 
institute a comparison, we may observe, that, if a 6-inch by 
12-inch yellow-pine header 6 feet 8 inches long, with four tail- 
beams 18 feet long, were to support a floor, the floor surface 
would be 96 square feet, giving 48 square feet to be supported 
by the header. This, if the floor were loaded with 100 lbs. per 
square foot, would bring upon the header 4800 lbs., or about 
one-half the breaking-weight of a header only 5 feet 4 inches 
long ; whereas, it would commonly be supposed, that, with such 
a construction for 100 lbs. per square foot of floor, we should 
have provided an unnecessarily large margin of safety. 

As to the fact that the header supported in stirrup irons 
bore less than that which was framed, this must be due to a 
difference in the quality of the timber ; and it would be unfair 
to conclude from only two tests that the second was a stronger 



544 



APPLIED MECHANICS. 



mode of construction than the first, even as far as the header 
itself is concerned. 

The fact, also, that a 6-inch by 1 2-inch yellow-pine beam 5 
feet 4 inches long bore 48000 lbs. centre load, equivalent to 
96000 distributed, without breaking, while the header broke at 
10916, shows what an enormous weakening is caused* by cutting 
mortises, and how much strength would be gained by avoiding 
all framing, and using stirrup irons to support the tail-beams in 
all cases where they cannot be supported on top of the header 
bearing the latter. 

No. 83. — Test of yellow-pine headers. 




Fig. 246. 



The headers, 6 inches by 12 inches, span 5 feet 4 inches, 
were hung at either end in iron stirrups, from trimmers 6 
inches by 12 inches by 20 inches, which in turn were sup- 
ported on jack-screws. The headers were mortised in three 
places (16 inches on centres) for three 3-inch by 12-inch yellow- 
pine tail-beams 10 feet in length. 

The load was applied at the centre of the tail-beams, and 
divided equally among the three by iron bridging. 

The tail-beams were cross-bridged in two places, 6 feet 
apart, by 2-inch by 3-inch spruce bridging, and also floored over 
with yellow-pine flooring 1 inch thick. 

The weight of the tail-beams, bridging, flooring, etc., which 
was supported by header, was 833 lbs. 

Weight of north header, 106 lbs. 

Weight of south header, 122 lbs. 



TRANSVERSE STRENGTH OF TIMBER. 545 

The headers were heart pieces, coarse grain, and sappy. 
The north one had a few season cracks extending from mortise 
to mortise on the outside. The tail-beams were of medium 
quality, two of them having sapwood on the edges; the third 
was much coarser, but contained more pitch, being heavy. 
Weight of yoke and iron bridging, 366 lbs. 

Details of the Test. 

15366 lbs. Cracks heard in north header. 

19866 " Loud cracks ; season cracks opening in north header. 

21366 " North header giving way, load dropped 500 lbs.; cracks 

§ inch wide. 
26366 " South header began to show cracks.^ 
26^66 " Centre tail-beam broke off below tenon at south end ; load 

dropped 1500 lbs. 

North header virtually broken at 21833 lbs., so each header 
bore 2 1 1 3 3 = 1 09 1 6 lbs. 

The tail-beam which did not break was heavy and full of 
pitch, and of a coarser grain than the other. 

No. 86. — Test of yellow-pine headers by means of tail-beams 
and floor. 

The headers, 6 inches by 12 inches, span 5 feet 4 inches, 
were mortised at each end into the trimmers with a double 
tenon and joint-bolt. The trimmers were supported on jack- 
screws, as before. Three taibbeams, 3 inches by 12 inches, 
span 10 feet, were mortised into the headers with tusk and 
tenon, and pinned. 

The load was applied at the centre of tail-beams, and dis- 
tributed equally over the three by means of bridging. (See 
No. 83.) Tail-beams cross-bridged in two places with 2-inch by 
3-inch spruce, and floored over with i-inch yellow-pine flooring. 

Weight of tail-beams, bridging, flooring, etc., which was 
supported by headers = 763 lbs. 

No deflections taken. 



546 APPLIED MECHANICS. 

Weight of yoke and iron bridging, 366 lbs. 

Details of the Test. 
14366 lbs. North header heard to crack internally. 
24366 " Two tail-pieces began to crack under lower tenon (north 

end). 
24866 " A few minutes later one of them broke under tenon (north 

end). The headers, as far as could be seen, were 

uninjured. 

No. 89. — Test of headers in floor (yellow pine). The 
headers and trimmers were the ones used in No. 86, and were 
framed in the same way. Three tail-beams, 3 inches by 12 
inches, 6 feet 6 inches span (inside measurement), were framed 
into headers with a tusk-and-tenon joint, and then pinned. 

The experiment was precisely the same as No. 86, with 
the exception, that, instead of 10-foot tail-beams being used, the 
length of these was 6 feet 6 inches. 

No deflections were taken. 

Details of the Test. 

* 20150 lbs. Season cracks in north header began to open wider. 

* 26325 " North header broke through the middle, following the 

line of mortises, then held 18825 lbs. 
South header cracked but little. 

No. 105. — Tests of yellow-pine header in floors. 

The headers and tail-beams were framed as in No. 86 ; 
and the experiment was exactly the same, with the exception 
that the tail-beams used were 6 feet Ions:. 

Details of the Test. 
20762 lbs. Season crack in both headers opened. 

23262 " North header failed, after three minutes, through the line of 
mortises ; while south header was but little cracked.' 

* This weight includes half the weight of bridging, floor, and tail beams 
(325 lbs). 



SHEARING OF TIMBER ALONG THE GRAIN. 547 

No. 106. — Test of yellow-pine header. 

In this experiment the headers were hung in stirrup irons, 
exactly as in No. 83. 

Details of the Test. 

24262 lbs. Slight cracking. 

26662 " East tail-beam split below line of tenon. 
30262 " Held for five minutes, when west stirrup on north tail-beam 
broke. The header was virtually broken. 

No. 10 J. — Test of yellow-pine header. 

The unbroken headers of Nos. 105 and 106 were used. 

Details of the Test. 

22262 lbs. Crack in north header. 

25162 " North header (framed) suddenly failed. 

§240. Shearing of Timber along the Grain. — The shear- 
ing of timber always takes place along, and not across, the 
grain ; for it can be shown, that, wherever we have a tendency 
to shear on a certain plane, there is an equal tendency to shear 
on a plane at right angles to it. Hence if there is, at any point 
in a piece of wood, a tendency to shear it across the grain, there 
must necessarily accompany it an equal tendency to shear it 
along the grain ; and, the resistance to the latter being very 
slight, the timber will give way in this manner, instead of across 
the grain. 

As to the shearing-strength per square inch, some values 
have been given in Rankine's table ; and the following table 
contains results obtained at the Watertown Arsenal, and re- 
corded in Executive Document No. 12, 47th Congress, first 
session. 



548 



APPLIED MECHANICS. 







Shearing- 






Shearing- 


Kind of Wood. 


Arsenal 


Strength 


Kind of Wood. 


Arsenal 


Strength 




No. 


per Square 
Inch. 




No. 


per Square 
Inch. 


Ash 


620 


600 


Oak (white) . . 


631 


752 




621 


592 


Pine (white) . . 


75 2 


324 




622 


458 




753 


267 




623 


700 




754 


352 


Birch (yellow) 


623 


563 




755 


366 




633 


815 


Pine (yellow) . . 


607 


399 




634 


672 




608 


3*7 




635 


612 




614 


409 


Maple (white) . . 


636 


647 




615 


4i5 




637 


537 




616 


409 




638 


367 




617 


3 6 4 




639 


43i 




618 


286 


Oak (red) . . . 


624 


775 




619 


330 




625 


743 


Spruce .... 


748 


253 




626 


999 




749 


374 




627 


726 




750 


347 


Oak (white) . . 


628 


966 




75i 


3i6 




629 


803 


Whitewood . . 


609 


406 




630 


846 




610 


382 



§ 241. General Remarks. — A perusal of the tests on 
columns and on beams will show that one of the principal 
sources of weakness in timber is the presence of knots, and it 
will be noticed that the position of the fracture is in most 
cases determined by the knots. 

Sap-wood, season cracks, and decay are doubtless other 
sources of weakness. The tests, however, do not present'such 
striking evidence of the deleterious effects of the first two as 
is the case with knots. In general, it may be said, however, 
that timber used in construction should be free, or nearly free, 
from sap-wood ; as an excessive amount of sap-wood renders it 
weak. 



GENERAL REMARKS. 549 



It will often be found to be a common opinion among lum- 
ber-dealers, that a piece of timber which contains the heart is 
not as good as one which is cut from the wood on one side of 
the heart. This is very often true ; as the timber which is sold 
in the market is very liable to have cracks at the heart, and 
also, if the tree has passed maturity, the heart is the place 
where decay is likely to begin. Nevertheless, the tests of 
beams would not, it seems to the author, bear out the conclu- 
sion that such pieces as contain the heart are always weaker 
than those that do not. 

Another matter that claims serious consideration is the 
effect of seasoning upon the strength of timber. This question 
can only be decided by tests on full-size pieces, as the small 
pieces season much more rapidly and uniformly than full-size 
pieces. 

In this regard, the observation should be made, that prac- 
tically our buildings and other constructions are built with 
green lumber ; i.e., lumber which has been cut from three 
months to a year. Unless it can be shown that the seasoning 
which the lumber receives while in use imparts to it a greater 
strength, it will only be proper to consider its strength the same 
as that of green lumber. Not very much evidence has thus far 
been obtained upon this point ; but, such as it is, it will be 
noted here. 

i°. We have, on p. 505, the results of the tests of a lot of 
old-mill columns ; and, while some of them did exhibit a greater 
strength than green ones, a perusal of this set of tests will 
convince the reader that it would not be safe to rely upon any 
greater strength in these columns than in green ones. More- 
over, these columns had been in a building heated by steam for 
a number of years, and during the seasoning process they had 
been subjected to the load they had to support. The writer 
has also observed some evidence of the same kind in con- 
nection with one of his time tests. 



550 APPLIED MECHANICS. 

2°. In the case of beams, we have, in Nos. 60 and 66, 
examples of beams which had been seasoning, unloaded, in a 
building heated by steam ; and in these cases there was a great 
gain in strength. Some yellow-pine beams exhibited a similar 
action. On the other hand, beams Nos. 18 and 19 had been 
seasoning on the wharf, in the open air, for about one year ; 
and while some yellow-pine beams which had seasoned without 
load, in the building, showed great strength, in other cases the 
increase was not so marked. 

In view of the fact that the above is practically all the evi- 
dence we have in the matter, it would seem to the writer, 
unless future experiments shall prove the contrary to be true, 
that we cannot rely, in our constructions, upon having any 
greater strength than that of the green lumber, and that the 
figures to be used should be those obtained by testing green 
lumber. 

§ 242. Strength of Building-Stones. — Inasmuch as it is 
not intended to enter into a discussion of the work that has 
thus far been done in testing the strength of the ordinary 
building-stones, and as it will be a convenience to the reader 
to have some figures on which he can depend with reasonable 
certainty, the following table will be given, taken from Gen. 
Gillmore's report. 

The specimens tested were cubes, 2 inches on a side : their 
faces were smoothed and polished, so as to obtain an even bear- 
ing. 

In the following table, B means "on bed," E means "on 
edge." 



STRENGTH OF BUILDING-STONES. 



551 



Kind. 



Gratiites : — 
Blue. . . , 

Dark . . , 
Light . . 

Flagging . 
Old Quarry , 

Up-river . 
Niantic River 

Harlem stone , 
Tombstone 

Porter's rock , 
Gray. . . , 



Locality. 



Staten Island, N.Y 

Fox Island, Me. 

Dix Island, Me 

Quincy, Mass 

Tarrytown, West Chester Co., N.Y. 

North River, N.Y 

Westerly, Washington County, R.I. 

Millstone Point, Conn. . . . , 

«< it it 

Sprucehead, Me 

Hewitt's Island, Me 

Richmond, Va 

Greenwich, Conn 

New London, Conn 

Fox Island, Me 

« it << 

Vinalhaven, Me 

Morrisania, West Chester Co., N.Y. 

Sharkey's Quarry, Me 

tt a ti 

Richmond, Va 

Cape Ann, Mass 

Mystic River, Conn 

Westerly, R.I 

Richmond, Va 

New Haven, Conn 

Stony Creek, Conn 

« « tt 
Fall River, Mass 



Strength 

per Square 

Inch. 



lbs. 
22250 

14875 
15000 
i775o 
i475o 
18250 
13425 
i775o 
17250 
16187 
18750 
13500 
17500 
14375 
15062 
21250 
20000 
11300 
11700 
12500 

I4I75 

15062 
11700 
13150 
16750 
15800 
22125 
20875 
16063 
12423 
19500 
18125 
22250 
14687 

14937 
14100 

13875 
7750 
9500 
15000 
16750 
ic575o 
15937 
9250 



Weight of 

One Cubic 

Foot. 



lbs. 
178.80 
164.10 
166.50 
166.20 
168.70 
162.20 
168 10 
165.60 
165.60 
168.70 
168.70 
171.90 
171.90 
164 60 
164.60 



177.20 
177.20 
166.25 
166.25 
166.30 
166.30 
170.00 
170.00 
170.00 
170.00 
170.00 
170.50 



164.40 
164.40 
166.90 
166.90 
164.40 
164.40 
162.50 
162.50 
165.40 
165.40 
165.40 
165.00 
i6t.oo 



552 



APPLIED MECHANICS. 



Kind. 



Granites : — 

Rose 

Gray 

Gneiss .... 
<< 

Gray 

Dark, soft-looking 
Gray 

<< 

a 

Limestones : — 
Glen's Fall- 



Lake . . 



North River 



White . 



Sandstones 
Brown . 

Gray . . 

Brown . 



Locality. 



Niantic, Conn 

Sachem's-head Quarry, Conn. 
Hurricane Island, Me 

Madison Avenue, N.Y 

Port Deposit, Md 

<< «< << 

Garrison's, North River, N.Y. 

Palmer Quarry, Me 

Rockport, Mass 

Glen's Falls, Warren County, N.Y. 

Lake Champlain, N.Y 

Canajoharie, Montgomery Co., N.Y. 

Kingston, N.Y 

Garrison's Station, N.Y. . . . 

<< (< << 

tl (f « 

Marblehead, O 

it ' tt 

« M 

Little Falls, N.Y 

Belleville, N.J 

Middletown, Conn 



Strength 

per Square 

Inch. 



lbs. 
9550 
945° 
15937 
14000 
14425 
14937 
1 1000 
1 1 250 
12500 
i975o 
13100 
16500 
15000 
12250 
13370 
1 1 500 
16500 
16300 
19750 



"475 
10750 
25000 
21500 
20700 
19250 
13900 
1 1 050 
18500 
18275 
17750 
i775o 
18775 
11250 
10700 
12600 



9850 
9^0 
1 1 700 
10250 

6950 



Weight of 

One Cubic 

Foot. 



lbs. 
162.50 
161.20 
163.70 
163.70 
166.90 
166.90 
166.90 
182 50 
182.50 
170.00 
170.00 
170.00 
170.00 
161.30 
161.30 
161.90 
161.90 
163.20 
163.20 



168.80 
168.80 
171.90 
171.90 
169.80 
169.80 
168.20 
168.20 
164.70 
164.70 
163.70 
163.50 
163.50 
150.00 
150.00 
i55-oo 



140.60 
140.60 
141.00 
141.00 

148.50 



CEMENT MORTAR. 



553 



Kind. 



Locality. 



Strength 

per Square 

Inch. 



Weight of 

One Cubic 

Foot. 



Sandstones 
Brown 
Red . 

Pink. 

Drab 



Middletown, Conn 

Haverstraw, N.Y 

Medina, N.Y 

Berea, O 



lbs. 
555o 
435o 
4025 
17250 
14812 
17725 
10250 
8300 
7250 



lbs. 
148.50 

133-10 
133-10 

150.60 

149-30 

151. 10 

131.90 
133.10 
137-50 



§ 243. Cement Mortar. — The reader who wishes to inves- 
tigate this subject is referred, for the beginning of his research, 
to the treatise of Gen. Q. A. Gillmore on " Limes, Hydraulic 
Cements, and Mortars." 

He made a large number of tests (and of these, only the 
following abridgment of one of his tables will be given) show- 
ing the resistance which Croton bricks cemented together cross- 
wise in pairs, face to face, offer to a force of traction applied at 
right angles to the surfaces of contact. 



Kind of Cement. 


Composition of Mortar, by Volume. 


Ultimate 

Tensile 

Strength 

per Sq. In. 


Delafield & Baxter 

«< << « 

(( K (« 

« u it 

Lawrence Cement Company 

tt tt (( 


Stiff paste of pure cement . . 

K H It U tt 

tt tt (< it it 

it it tt (< tt 

a u a ti it 
u « a it <c 
it « << it i« 


IO97 
IIOI 

( less than 
( 420 

843 

898 
1258 
1242 



554 



APPLIED MECHANICS. 







Ultimate 


Kind of Cement. 


Composition of Mortar, by Volume. 


Tensile 
Strength 
per Sq. In. 


Lawrence Cement Company 


St 


iff paste of pure cement . . 


1284 


«« <« <« 






< < 








1398 


Kingston & Rosendale . . . 






< < 








1227 


« « tt 

<( (( <( 










« < 
< < 








969 
836 


«( (( tt 










( « 






. . 


I284 


«« (( «« 










< c 








I055 


Hancock, Maryland 










' ' 








648 


" " 












' ' 








777 


« it 












i 1 








1023 


it « 












1 < 








617 


Newark & Rosendale 












' ' 








1213 


James River . . . 










1 a < 








859 


Delafield & Baxter 










Cement in powder, 4 ; sand, 1 . 


1023 




< <( <« 
( it << 
( « (« 










« « 

« « 




1420 

763 
1023 




( it . it 










2 


a « 


1030 




( C( «( 










<( « 


<« « 


1113 




i tt «( 










<« « 


«( tt 


1070 




< << <( 










« « 


11 u 


'843 
420 




( << <« 
< «( « 










« «« 

«« « 
<« « 


« <( 
«( «< 


73 2 
812 

523 




< «« «« 










<« « 


" 2 . 


367 




< (< << 










K «< 


«< « 


260 




« «« « 










«< u 


u « 


491 


James River . . 










8 


" I . 
tt << 


978 
812 


U tt 










4 


" " . 


992 


a it 










« . « 


M (( 


300 


a tt 










« 1 


<« C( 


740 


tt tt 










« « « 2 . 


392 



FUNDAMENTAL PRINCIPLES. 555 



CHAPTER VIII. 
CONTINUOUS GIRDERS. 

§ 244. Fundamental Principles. — A continuous girder is 
one that is continuous over one or more supports ; i.e., one that 
has at least one support in addition to those at the ends. The 
principle of continuity is, that the neutral line is throughout a 
continuous curve over the supports, the tangent to one branch 
of the curve at the support being a prolongation of the tangent 
to the other branch. 

Whereas, in the girder supported at the ends, the bending- 
moment at the support is zero, in the continuous girder there 
is a bending-moment at the support, where the girder is con- 
tinuous. There is also a shearing-force at each side of the 
support, the sum of the shearing-forces on the two sides of 
any one support forming the supporting-force. 

In this chapter will be given the general methods of deter- 
mining the bending-moments, slopes, and deflections of con- 
tinuous girders. 

i°. When the loads are distributed. 

2°. When the loads are all concentrated. 

3 . When there are both distributed and concentrated loads. 

It is believed that the reader will thus have the means of 
solving all cases of continuous girders, and that, whenever it 
is desirable to have a set of simplified formulae for a small but 



556 



APPLIED MECHANICS. 



definite number of spans, or for some special proportions or 
distribution of the load, he will be able to deduce such simpli- 
fied formulae from the more general ones. 

§ 245. Distributed Loads. — In this case we assume that 
all the loads are distributed, whether they are uniformly dis- 
tributed or not. The first step to be taken is, to find the bend- 
ing-moment over each support : this is done by using what is 
known as the " three -mom e7it equation" which we shall now 
proceed to deduce ; and, in the course of the reasoning by which 
we deduce it, we shall derive a number of useful equations, ex- 
pressing bending-moment, shearing-force, slope, deflection, etc., 
at various points. 




Fig. 247. 



For the purpose in view, let us assume our origin at O 
(Fig. 247), and let 

M t — bending-moment at B. . 

M 2 = bending-moment at O. 

M 3 — bending-moment at A. 

/, = OA. 

/_, = OB. 

F Q — shearing-force just to the right of O. 

F_ Q = shearing-force just to the left of O. , 

F t — shearing-force at distance x to the right of origin. 

F_ x = shearing-force at distance x to the left of origin. 

Shear is taken as positive when the tendency is to slide the 
part remote from the origin upwards. 

If S = supporting-force at O, 



F Q + F_ 



DISTRIBUTED LOADS. 557 

Beginning, now, by taking O as origin, and x positive to the 
right, — 

Let OC = x. 

CD == v = deflection at distance x from origin. 
w = load per unit of length (either constant, or vari- 
able with x). 
We shall then have, from the principles of the common 
theory of beams, 



F, = F Q 



■ I wdx; (i) 



i.e., the shearing-force at a distance x to the right of O is 
found by subtracting from the shearing-force just to the right 
of O the sum of the loads between the section at x and the 
support ; and this sum is 



f 



wdx. 



In a similar manner, if we were to take origin at 0, and x 
positive to the left, we should have 



F_t = F_ 



I wdx, (2) 



In § 204 we found the equation 

dM = F 
dx 



• —r- = F — I wdx. 
dx J 



Hence, integrating between x = o and x = x, and observing, 
that, when x = o, M = 7fcT 2 , we have 



M — M 2 = Fox — j f wdx 2 , 

I/O I/O 



558 APPLIED MECHANICS. 

which reduces to 

M= M 2 + FoX - J J wdx 2 ; (3) 

Jo *J Q 

or, in words, — 

The bending-moment at a distance x to the right of is 
equal to the bending-moment over the support at the origin, 
plus the product of the shearing-force just to the right of the 
origin by the distance of the section from the origin, minus 
the sum of the moments of the loads between the section and 
the support about the section. 

Observe that this sum of the moments of the loads between 
the section and the support about the section has, for. its math- 
ematical equivalent, the expression 



%) o ISO 



wdx 2 ; 



and, as a particular instance, it may be noted, that when the 
load is uniformly distributed, and hence w is constant, this will 
reduce to 

WX Z , . X 

= (wx) -, 

2 2 

wx being the load between the section and the support, and - 

being the leverage of its resultant. 
Now write, for brevity, 



1/0 (/O 



wdx 2 = m; 

1/0 t/o 

then 

M = M 2 + F x - m. (4) 

Now, from § 194, we have 

d 2 v __ M 
dx 2 EI' 



DISTRIBUTED LOADS. 559 

Let a x = slope at distance x to the right of the origin. 
a_ x = slope at distance x to the left of the origin, 
a,, = value of a t when .r = o. 
a_ == value of a_ z when x = o. 
Then 



tana x = — = I — -ax 
dx J EI 



+ c, 



where c is an arbitrary constant, to be determined from the 
conditions of the problem. 

If, now, we substitute for M its value M 2 + F x — m, we 
shall have 

dv **■ C* dx z-, C x xdx f x mdx , 



To determine c> observe, that, when x =z o, a x = a ; 

.*. c — tana 

.-. tana s = £=tanao + ^r^ 

, zp C* Xdx f* x mdx , x 

Integrate again, and observe, that, when x = o, v = o, and we 
obtain 

v = xtzna + M 2 f C 4?L 

w r x r x xdx* r x r x mdx^ , A . 

+ H J. ^r - i X -bT- (6) 



560 APPLIED MECHANICS. 



Now write, for the sake of brevity, 

JftX pX pj pX pl— x px 

I wdx 2 , m 1 = I I wdx 2 , m_ z — I I wdx 2 , 

O Jo Uo Uo *J fJo 

r x r x dx 2 r h r^ 2 C lx C x d& 

r x r x xdx 2 c h r x xdx* r l ~ i r x xdx* 

y-J.i.^r *-ll ?; '-"./. Jo */• 

^-J a l EI' K '-J J EI' F --J J EI' 



the last four being derived by taking x positive to the left. 
We shall have 

v = xt3ina + M 2 n + F Q q — V; (7) 

and, if v x = deflection at A = vertical height of y4 above 0, we 
shall have, by substituting /, for x in (7), 

z;, = i l tan a Q 4- -^2^1 + B Q q x — ^i- 

Now, if we assume any horizontal datum line entirely below all 
the points of support, and let the height of B above this line be 
y bi that of A, y a% and that of 0, y of etc., we shall have 

y a - y = h tan a Q -t- M 2 n, + F Q q x - V % . (8) 

And, if we put x = / x in (4), we shall have 

^T 3 = M 2 + ^ / x - m s 

, ,.-">- f + m ". . ( 9 ) 

and, if we substitute this value" of F Q in (8), we obtain, by redu- 
cing, 

/ tfA M,q I m l q l 

y a -y Q = /xtanoo + M 2 \n t - jj + -f- + -~ - ^ij 



DISTRIBUTED LOADS. 



5 6l 



and, solving for tan a , we obtain 



tan a n = 






+ M S 



«" 



- 1 _ MA^ - 



Vr 



m t q t 



+ T (lo) 



This expression gives us the tangent of the slope at O in span 
OA ; and equation (9) gives us the shearing-force just to the right 
of O in span OA, in terms of M 2 , M 3 , and known quantities. 

If we were to take the origin at O, as before, and x positive 
to the left instead of the right, we should have, in place of (4), 



M 



in place of (9), 
and in place of (10), 

yb — y a 



tana_ Q = 



/-x 



F_ t 



+ M 2 



M 2 + F_ Q x — m; 
M x — M 2 -f- m_ l 

V-x 2 I- J 



(12) 



q- 



V-i 



(13) 



But, since the girder is continuous, we must have the tangent at 
O to the left-hand part, a prolongation of the tangent at O to 
the right-hand part, as shown in Fig. 248. 
Hence we must have 



a_ = — a Q 
tana_ -f- tana = o. 



Hence, adding (10) and (13), we 
have 




y a 



yo n 



0r+">{% 



- M* 



z h 2 



M> 



m l q 1 



**-x?-x K V-r 

/-x 2 + /x + /-x 



(14) 



and this is the "three-moment equation" for the case of a dis- 
tributed load, whether it be uniformly distributed or otherwise. 



562 APPLIED MECHANICS. 



CASE WHEN SUPPORTS ARE ON THE SAME LEVEL. 

When the supports are all on the same level, then y a = y b 
y oy and the three-moment equation becomes 



^1 ^1 q~i «_i) M z q l M x q_ x 






/x 2 



- ^ - "T^T + 7i + j- = °- (*5) 

MANNER OF USTNG THE THREE-MOMENT EQUATION. 

When the dimensions and load of the girder are known, all 
the quantities in the three-moment equation, whether we use 
(14) or (15), are known, except the three bending-moments, M iy 

Suppose, now, the girder to have any number of (say, seven) 
points of support ; then, by taking the origin at B (Fig. 247), 
we obtain one equation between the bending-moments at E> B, 
and O, the first of which, if is is an end support, is zero. Next 
take the origin at O, and we obtain one equation between the 
three bending-moments at B, O, and A; and so, continuing, we 
obtain five equations between five unknown quantities. 

Solving these, we obtain the bending-moments over the 
supports ; and from these bending-moments, after they are 
found, we can obtain the shearing-forces, bending-moments, 
slopes, and deflections, by using the equations deduced in the 
course of the reasoning for the three-moment equation, as equa- 
tions (4), (5), (7), (9), and (10). 

SPECIAL CASE, 

when, the supports being all on the same level, the load on any 
one span is uniformly distributed over that span, and when the 
girder is of uniform section throughout. 



DISTRIBUTED LOADS. 563 

Let w x — load per unit of length on span OA, origin at O. 
w_ x = load per unit of length on span OB, origin at O. 
I = the constant moment of inertia of the section 

y a = yb — y<y 



Then 



w x l x 2 

mi = , 

2 

/x 2 

n ^^Ef 
qx " 6Ef 



tn_ x 


= 


2 


z 


w_i 


= 


2EI' 




?-i 


= 


l_j> m 

6EI' 




F . 




w_ x l_ 


c 4 



24EI 24EI 

With these substitutions, the three-moment equation, either 
(14) or (15), becomes 

M x l_ x + 2M 2 (/_ X + /i) 4- i^/r + \{wj? 4- w_J-*) = o. (16). 

This is a simpler form of the three-moment equation, applicable 
to this particular case only. 

Example I. — Suppose we have a continuous girder of 
uniform section, uni- 
formly loaded, and a t-tt- *- a 
of three equal spans, l ' 49 ' 
to find M B and M c , also the supporting-forces, shearing-forces, 
bending-moments, slopes, and deflections throughout. 

Solution. — Take the orisrin at B, and we have 



M x = 0, 




M 2 = M z = M B = M c ; 


since 








h 


= u = * 


equation (16) gives 






5 MJ + \wl* = 





tut *r w l 2 

,\ M B = M c = 

10 



564 APPLIED MECHANICS. 

Next, to find the shearing-forces, we have, from (9), 

wl 2 wl 2 wl 2 

10 io 2 ' wl 

jt b = A_ c = = — > 

I 2 

equals shearing-force just to the right of B or left of C. 
Shearing-force just to the right of C or left of B = 

wl 2 wl 2 

IO 2 „ - 

- = \wL 

Hence supporting-forces are 

S B = S c = (f + \)wl = J>/, 

S K = ^ D = J (30// - fjze//) = \wl 

Bending-moment in span AB at distance x from ^4, or in 
span CD at distance x from Z>, 

**• o ? wx2 
M = %w/x — . 

5 2 

Bending-moment in middle span at a distance x from B or 

from £7, 

-, wl 2 , «//# m* 2 

10 2 2 

Shearing-force in span AB or CD at a distance 4r from A 
ox D, 

J? = ^wl — wx. 

Shearing-force in middle span at distance x from B or C, 

F — — • — wx. 
2 

Maximum bending-moment in span BC (when x = -), 

-- wl 2 , ze// 2 a// 2 wl 2 

M Q = - — + — - — = . 

10 4 5 40 



DISTRIBUTED LOADS. 565 

Maximum bending-moment in span AB or CD, 

* = ¥> 

Mo = ±wl 2 - ±wfi = =£. 
25 50 25 

Hence the greatest bending-moment to which the girder is 

subjected is that at B or C, and its amount is — . 

10 

Slope at B in middle span, from equation (10), 

_wl 2 / I_\_L_ wl2 l l \_ wlz w l z 

ana B - — I jEj) + —\fi£j) ^EI + ^EI 

_ wl 3 / 1 1 1 \ _ wl* 

EI\2p 60 24/ 120EI 1 

which denotes an upward slope at B towards the right. In the 
same way, the girder slopes upwards at C towards the left. 
The slopes at B and C in the end spans are, of course, down- 
wards. 

Slope in the middle span at a distance x from B, 



dv 1 ( wl 2 . wlx 2 
tan a = — = — < 
dx EI\ 



1 ( wl 2 , wlx 2 wx 3 ) , 
EI( 10 4 6 ) 

When x — o, 



wl* wl* 

tana = H /. c = 



\oEI 120EI 



W ( /3 l 2 X , Z* 2 #3) 

tan a = — ^ 1 7 | 



EI (120 10 4 6 



(/3 — I2/ 2 JP + 30/^ — 20tf 3 ) 



I20EI 



W 

Deflection = v — ttJJ 1 x — 6l 2 x 2 -J- iolx* — kx*). 

1 2oE7 



566 APPLIED MECHANICS. 

In order to make plain all methods of proceeding, the slope 
in the end spans will be found in two different ways, as 
follows : — 

For bending-moment, slope, and deflection in left-hand span 

at a distance x from B '(or in the right-hand span at distance x 

from C), we have 

*jt W I 2 , 3 7 wx * 
M = h -wlx . 

io 5 2 

dv i ( wl 2 x , -zw/x 2 wxi) . 

tan a = — = — \ j- £ — \ -he. 

dx EI{ io io 6 ) 

When x — o, 



tan a 



tana = — 


120 EI 


I20EI 


dv w ( 
Tx ~ EI{ 


11 I 2 * , 3 7 , 

120 IO IO 


-?} 



.(— / 3 — I2/ 2 .# -f- 36/^ — 20* 3 ) 



I20EI 



Deflection = v == —(—A* — 6/ 2 ^ 2 -f- 12/^3 — 5*4). 

120 EI 



We may, on the other hand, accomplish the same object by 
finding the slope and deflection in left-hand span at distance 
x from A, or in right-hand span at distance x from D, as 
follows : — 

dx EIJ \s 2. J M 5 6 J 

When x = /, 

w/ 3 
tan< 



120EI 



wl*> Wl 3 , . _ W/ 3 



I20EI 30EI 40EI 



DISTRIBUTED LOADS. 567 

dv __ w { / 3 . lx 2 _ x 3 ) 

dx El\ 40 5 6 j 

— 3/3 + 24/jc 2 — 20^3 \ 



12 



oEI\ 



The figure shows the mode of bending of the girder. 



Fig. 250. 

To find the greatest deflection in either span, put the ex- 
pression for the slope equal to zero, and find x by the ordinary 
methods for solving an equation of the third degree, and then 
substitute this value in the expression for the deflection. 

Example II. — Continuous girder of two equal spans, sec- 
tion uniform, and load uniformly dis- 

' J A B C 

tnbuted. A * * 

Solution. — Take origin at B. 

M^ = M z = M A — M c = o, M 2 = M B , /, = / 2 — l } w 1 = w 2 = w; 
therefore, from equation (16), 

4 M B / + W = .'. M* = - — . 



Shearing-force either side of B 



wl 2 wl 2 

j — = W- 



Supporting-force at B = \wl. 
Supporting-force at A and C = \wl. 
Shear at distance x from A or C. 



F = \wl — wx. 



568 



APPLIED MECHANICS. 



Bending-moment at distance x from A or C, 
_ 3^ , wx 2 



M= -wlx 
8 



Maximum bending-moment occurs when x = f/, 

gw/ 2 



M = -^-ze// 2 2-ze// 2 



128 



12, 



Hence greatest bending-moment to which the girder is sub- 

tvl 2 

jected is that at B, and its magnitude is . 

8 

Slope at B, from equation (6), 
tana B = tana_ B == -_(__J___ + __ 

zsY (24 12 24 ) 
as was to be expected. 

Slope at distance x from A in span AB, 



tana 



/i*&- - ^\ 



EI \i6 



+ '• 



When x =. I, a = o ; 



C= - 



Z#/ 3 



48^/ 

tana = * = ^ U/* 2 - £ - Zii 
<2* ^/(i6 6 48) 



z» 



Deflection, 



48^/ 



(glx 2 - 8^3 _ /s). 



v = 



Z0 



(— A# + 3/^ _ 2*4). 



48EI 

For maximum deflection, we have 

6 -/x 2 = o 



16 



6 48 

# = 0.44/. 



DISTRIBUTED LOADS. 569 

Maximum deflection = — — \ — 1 + 3(0.44)2 — 2(0.44)3 ( (0.44) 
48^/ ( ) 

= — 0.0054 

^EI 

Example III. — In order to solve a case where no simplifi- 
cations enter, on account of symmetry or otherwise, we will 
take a continuous girder of five spans (as shown in the figure), 
the spans varying in length from 3/ to jl ; the loads being 
uniformly distributed, and varying in intensity from yuu on the 
longest span to jw on the shortest ; the beam being of uniform 
section. 

A 31 B 41 C 51 ID 6Z E "d F 

A ~w A Gw A bw A iw A Zw A 

Fig. 252. 

For this case we can use equation (9). 
Origin at B, 

o + i 4 /M B + 4 IM C + J[7zK27/ 3 ) + 60/(64^)] = o, 
or 

$6M B + i6M c = -573^/2. 

Origin at C, 

4 IM B + iMM c + 5 /Jf D + ^[6(64) + 5(125)] = o, 
or 

i6M B + 72^ + 2oM D = —loogwl 2 . 

Origin at D, 

5/J/c + 22/J/ D + 6/M s + £[5(125) + 4 (2l6)]^/3 = o, 
or 

2oM c + 88M D + 2 4 M K = —1489a// 2 . 

Origin at E, 

7P)/3 

6Df D + 2 6/M K + —[4(216) + 3(343)] = o, 
4 
or 

2 4 M D -f io 4 M E = — iSg^w/ 2 . 



57o 



APPLIED MECHANICS. 



The four equations are : 

S6M B + i6M c = - 573a// 2 . (i) 

i6M B 4 72i¥" c 4- 2oJI/" D = — 1009a// 2 . (2) 

2oJ/" c + 88^f D -f 24M E = —1489a// 2 . (3) 



24M» + i04J/ E = -1893a// 2 . (4) 



Eliminate M E between (3) and (4), and we obtain 
130J/C -f 536^0 = — 6839a// 2 . 



(5) 



Eliminate M D between (2) and (5), and we obtain 
2144^/3 + SggSM c = — ioiono// 2 . 



(6) 



Eliminate M c between (1) and (6), and we obtain 

234792^ = — 1769839a// 2 . 

.-. • M B = — 7-5379 w/2 > 

.'. from (1), M c — — 9.4299a// 2 , 

from (5), M v = — 104722a// 2 , 

from (4), M E = -15.7853a// 2 . 



(7) 



Shearing-force just to the right of 
o - 7-3579 + 3 T -5 



A = 



wl 



= 7.9874a//, 



„ —9.4200 4- 7-S37Q + 48 , 
^ = ^ yvT/3>);y ^ ^/ = 11.5270a//, 

— IO.4722 + Q-42Q0 + 62.5 

C = — y ^ -a// = 12.2915a//, 



_ —1=5.78=53 + 10.4722 4- 72 • 
2} == <■ — — —wl ~ 11.1145a//, 



^ = 



15-7853 + 73-5 



wl 



= 12.7550a//. 



DISTRIBUTED LOADS. 571 

Shearing-force to the left of 

n 7-5379 + 3i-5 , , j 

3 6 

-7-5379 + 9429 9 + 48 
C = 26// = 12.473026//, 

-9.4299 + 104722 + 62.5 
Z> = 26// = 12.70842c//, 

,-, -1 0.4722 + 15-7853 + 72 QQ 

7s = 7 a// = 1 2. 885526//, 

j, -i5'7853 + 735 , Q - 

F = 26// = 8.245026//. 

Supporting-force at 

^ — 7-987426//, C == 24.764526//, E = 25.640526//, 

i? = 24.539626//, D — 23.822926//, 7? = 8.245026//. 

Shearing-force at distance x to the right of 

A in section AB = 'j.gS'j^w/ — ywx, 

B in section BC = 11.527026// — 626/^, 

C in section CZ? = 12.291526// — $wx, 

D in section DE = 11.1 14526// — 426/^, 

E in section T^ = 12.755026// — 326/^. 

Bending-moment at distance x from 

726/jic 2 



^4 in section AB = -f- 7.987426/Zv — 

.5 in section BC = — 7. 537926/^ -f 11.527026//X: — 
C in section CD = — 9429926/Z 2 + 12.291526/Zx: — 
Z> in section DE = —10. 472226/^ -j- 11.114526/Z*: — 
is in section TsT' 1 == — 15.785326// 2 -f 12.755026//^ — 



2 ' 

676/X 2 

526/ 1 %' 2 

4wx 2 
~7~' 
326/jc 2 



572 



APPLIED MECHANICS. 



For the sections of maximum bending-moments (put shear- 
ing-force = o), — 

In AB, x = 1.1410// 
In BC, x = 1.9211// 
In CD, x = 2.4583// 
In DE, x = 2.7786// 
In EF, x = 4.2517/. 



in 



Hence the maximum bending-moments are respectively, 

Section AB, 

wxi-— + 7.9874/^= +4.5570W/ 2 . 

Section BC, 

— 7-5379^ /2 + 0/^(11.5270/— 3*) = 3.5347a// 2 . 
Section CD, 

— 9.4299W/ 2 -f- 0/^(12.2915/ — §.*) = 5.6781a// 2 . 
Section Z?is, 

— 10.4722a// 2 + 0/^(11.1145/— 2jr) = 4.9693a// 2 . . 
Section i£F, 

— 15.7853a// 2 4- 0/^(12.7550 — fjc) = 11.3297a// 2 . 



Values of tan a = slope in every case in the span, towards 



the right. 



tan 






Slope at B, 



a// 3 / 2 



)/2\a//3 32a// 3 16a// 3 



/! 



= O.337I 



a// 3 

jsz" 



DISTRIBUTED LOADS. 



573 



Slope at C, 

W l l fS S\ (S\wl l 625 7£//3 

625 wl 3 _ wl 3 



Slope at D, 



tana D 



wl 3 



-10.4722-^(1 - 3) + 15.7853^7 " 



24 ^Z^- 1 - 5983 ^/ 

7£// 3 J2Wl 3 $6wl 3 



EI 



+ 



EI 
0.7297 



EI' 



Slope at is, 

^ 3 /7 7\ 1029 7<y/ 3 343 w/ 3 
tan a E = - 15-7853^7^ - " j - -7T 77 + 



wl 3 

= -6.0426^- 



The manner of bending, very much exaggerated, is shown in 
the accompanying figure. 



Fig. 253. 

Slope at A = -4.096^, slope at F = +23.4578^-. 

For the deduction, see what follows. 
Slopes in General. 

Span AB, origin at A, 



tana 



ay ( 7 ) 

^7 j 3.9937^ -^ 3 [ 



+ '• 



When * = 3/, tan a = 0.3371^ ; 

U)l 3 f v . 7C>/ 3 

^7^35-9433 ~ 3i-5) + ' = °'337i^7 
.a// 3 



w i 7 ) 

tana = ^7<3-9937^ 2 - £* 3 - 4-io6/ 3 j 



574 



APPLIED MECHANICS. 



Span BC, origin at B, 



wl 1 



wl 2 



wlx 2 wx l 



tana = 0.3371— - 7.5379-gjr + 5-7635 £ 



w \ 5 

tana = ^=r- 7 .< — 1.5983/3 — 94299/ 2 # + 6.1458/* 2 — -x* 



Span CD, origin at C, 

Span DE, origin at D, 
tana = Jj J 0.7297/3 - 10.4722A* -f 5.55725/^ - -^sl. 

Span .fiT 7 , origin at E, 

tana = jU -6.0426/3 - 15.7853/ 2 * + 6.3775/^ - w -^-\. 

When x — 7/, 

tana = —(-6.0426 - 110.4971 + 312.4975 - 172.5) 

JtLl 



Deflections. 
Span AB, 



. Z£// 3 



ze/ ( 7 ) 

z; = -^jj 1.3312/^3 - —x* - 4.io6/3*|. 

Span BC, 

v — — J0.3371A3; — 3. 7689/2^ -f 1.9211/^3 — — I. 

Span CD, 

w ( 5 ) 

v = jj \ -1.5983/3^ - 4.7149/ 2 * 2 -j- 2.0486/^3 - — x *l 

Span DE, 

v = — J0.7297A* — 5.236i/ 2 ^ 2 + 1.8524 + Ix z — — i- 
is/ ( o ) 

Span EF, 

v = ^L i -6.0426/3^ - 7.8927/^ + 2.1258/^3 - £![. 
iiY ( 8 ) 



< > < 

xn-l n Xn 



n + 1 



CONTINUOUS GIRDER WITH CONCENTRATED LOADS. 575 

The maximum deflections can be obtained by putting the 
slopes eqital to zero, as before. 

§ 246. Continuous Girder with Concentrated Loads. — 
For our next general case, we will 
take that where there are no dis- — ^ — l 
tributed loads, but where all the n ~ 1 
loads are concentrated at single 
points, and the section uniform Wb "1, Wn 

r Fig. 254. 

throughout ; and we will begin by 

assuming only one concentrated load on eacji span: 

Let the support marked n — 1 be the (n — i) th support, and 
the length of the (n — i) th span be /*__,; let the load on this 
span be W n _„ and likewise for the other spans. Assume the 
origin at n> and let 

F„ = shearing-force just to the right of n. 
F_ n = shearing-force just to the left of n. 
F l = shearing-force at distance x to the right of n. 
F_ , = shearing-force at distance x to the left of n. 
Shear is taken as positive when the tendency is to slide the 
part remote from the origin upwards. 
If S n = supporting-force at n t 

S n = F n + F_ n . (1) 

Let, also, x n = distance from origin to point of application 
of load W m and let x n _ x = distance from origin to point of 
application of load W n _ I . 

Take x positive to the right. Then, for 

x < x H} Ft = F n ; ) * . 

x > x n , F t = F n - W n .\ K } 



Moreover, we have 



dx 



57^ APPLIED MECHANICS. 

hence, by integration, for 

x<x n , I dM = / F n dx; 

JM m Jo 

Jf*M f*x f*x 

dM = I F n dx— I W n dx + c; 
M n Jo Jo 

the value of c being determined from the condition, that, when 
x = x„ the two results must be identical. Hence we have, for 



x < x n , M = M n + F„x; 

x>x ni M= M n + F n x - W n (x - x„). 



(3) 



Make x = l n in the last equation, and we have 

M u+ x = M n + FJn - W H (Jn - X n ). ( 4 ) 

Now let l n — x„ ■= a n , and (4) becomes 

Mn + t = M H + FJn - Wndn', (5) 



hence 



p = Mn + I ~ Mn + ***** (6) 



Moreover, we have, as before, 

dx 2 EI dx 2 . 

/ being a constant. 
Let, as before, — 

a, = slope at distance x to the right of origin. 
a_ x = slope at distance x to the left of origin. 
a n z=z value of a x when x = o. 
a_ w = value of a_ x when x = o. 



CONTINUOUS GIRDER WITH CONCENTRATED LOADS. 577 

Then by integration, determining the constant in the same 
way as in (3), we have, for 

x 2 
x < x n , i?/(tan aj — tan a n ) = M n x + F n — ; 

2 



x > x„, ^/(tan a, - tan a„) = M n x + F„- - 



K7) 



2 2 

Hence 

x < x n , EI— — EI tan a n + M n x -f F n — ; * 

dx 2 

^jdv UT ^ . lir . ^x 2 W n (x — x n ) 2 

x > x m EI— = EI tan a n -f M n x -f F„ — ^ ^—. 

dx 22 

Integrate again, and determine constants in the same way, 

and for 

x 2 x 3 

x < x n , EIv = EIx tan a n + M n f- F n — : 

2 6 



* > *„, .£/* = EIxtena n + ^f„- + £„- - -?*^ ^ 

2 6 6 



(8) 



Make x = l n in the last equation, and denote the heights 
of the supports above the datum line in the same way as in 
§ 245, and we have 

EJ(yn + z—yn) = FIl n tan a n 

+ Jf tf + ^ - *(»-«.)' . (9) 

26 6 

Substitute for l n — .*•„, <z„, and for F m its value from (6), and we 

have 

EI(}'n + 1— y n ) = £74 tan a« 

+ ifi^L + J£ + 1 £ + i^(4 2 _ «..). (10) 
3 6 6 

Hence 

£/tana w + ^(2M n + M n + 1 ) + -^(/^ - *„*) 
6 64 

= jE/ Ow^Z»). (II) 

• in 



578 APPLIED MECHANICS.. 

Now, if we take origin at n and x positive to the left, we 
should obtain, instead of (n), 

EI tan a_ n + l ^{2M n + M n _ z ) 
6 

+ Wn -,•*"-* (4 -,' - a n _ >) = M*i£±*\ ( x 2 ) 

64-1 \ 4_i / 

Now add (n) and (12), and observe, that, since the girder is 
continuous, 

tan a n + tan a_ n = o, 

and we obtain 

M I M 7 

-^(L + 4_z) + M n ^!^ + ^±^ 
3 6 6 

64 64 _ x 

= ^/) - > '^ ■-■>'" + ->'"-■ ->'" [; (13) 



and this is the "three-moment equation" for the case of a 
single concentrated load on each span, and a uniform section. 
When the supports are all on the same level, this becomes 

^"(4 + 4_.) + X..k? + %^ + ^(4* - «»*) 

3 6 6 64 

+ ^-/^- I (4- I 2 --^-z 2 ) = o. (14) 

Either of these equations can be used (when it is appli- 
cable) just as the three-moment equation was used in the case 
of distributed loads. 






CASE OF MORE THAN ONE LOAD ON EACH SPAN. $?g 



CASE OF MORE THAN ONE LOAD ON EACH SPAN. 

When there is more than one load on each span, the three- 
moment equation becomes as follows : — 

^(4 + 4_0 + M n J-^fl + M n + ^ 
3 6 6 

+ 2^(4 2 - a*) + S^~ I ^- I (4- I 2 - a n _f) 
64 64_! 



' 4 4 — i ) 



In using these equations for concentrated loads, we can 
determine the moments over the supports ; but we must observe, 
that, in getting slopes and deflections, bending-moments, etc., 
the algebraic expressions that represent them are different on 
the two sides of any one load, and hence we must deduce new- 
values whenever we pass a load, determining 'the constants for 
our integration to correspond. 

Example. — Given a continuous girder of three spans, the 
middle span = 20 feet, each end span = 15 feet; supports on 
same level. The only loads on the girder are two ; viz., a load 
of 5000 lbs. at 5 feet from the left-hand end, and one of 4000 
lbs. 5 feet from the right-hand end. The supports are lettered 
from left to right, A, B, C, D, respectively. Find the greatest 
bending-moment and greatest deflection. 

Solution. — Origin at B> 

_J?( 20 + I5 ) + -^(20) + * 5 (22 5 - 2 5 ) = O. (l) 

3 6 o X 15 

Origin at C, 

M Z( . v . M B , v , 4000 x 5, N , \ 

— (20 + 15) + —.20) + *-— -(225 - 25) = o. (2) 

3 6 6 X 15 



58o 



APPLIED MECHANICS. 



These reduce to 








70 M B + 20M C + 


I OOOOOO 

3 


= 


M B 


= —4000 foot-lbs. 


2oM B -f- joM c -f 


800000 
3 


= 




= —2667 foot-lbs. 


Shearing-forces. 




Supporting-forces. 


Slopes at supports. 


^a = 3 o6 7> F_ c = 


-67. 


S A = 3067. 


tan a A = - 59444. 
EI 


E_ B = 1934, ^ c = 


1511. 


S B = 2000. 


tana B = -f 35557 . 
EI 


^ B = 67, ^_ D = 


2489. 


S c = 1444. 


tiiii 
tan a c = — — -. 
EI 






S D = 


= 2489. 


48888 
tan a_ D = — - 



EI 

Span AB y origin at A, 

x < 5, M — 3067^. 

x> 5, J/= 3067* - 5000O - 5) = 25000 - 1933*. 

Maximum bending-moment occurs when x = 5 and therefore 

M Q = 15333. 

# < 5, £7 tan a, = -59444 + I533* 2 '■> 
x > 5, EI tan a x = 25000.* — 967^ 4- *". 

Determine ^ by condition, that, when x — 5, these two become 
equal ; 

.*. r = —1 21944; 

/. x > 5, is/tan aj == — 121944 -f- 25000* — 967^. 

For deflections, 

# < 5, .£Yz/ = — 50444.2; -f- 511.* 3 ; 

x > 5, .£yz> = —121944.* + 12500* 2 — 322^ -f- *• 

Determine £ from condition, that, when x = 15, z/ = o; 

\ <r = 103410; 

.*. a: > 5, isT^ = 103410 — 121944.* -j- 12500^ — 322JC 3 . 






CASE OF MORE THAN ONE LOAD ON EACH SPAN. 58 1 

For maximum deflection, equate slope to zero, and find x. 
We find it at x = 6.53. 
.-. EIv = —249531. 

Span BC, origin at B, 

M = —4000 -f- 67X, 
EI tan a, = 35557 - 4000* + 33X 2 , 
EIv = 3555 7# — 2000JC 2 + ir# 3 . 

For maximum deflection, equate slope to zero, and find x. 

We find it at x — 9.78. 
.*. EIv — 166740. 

Span CD, origin at C, 

x < 10, M — —2667 + 1511AV 

x > 10, M = —2667 4- 1511^ — 4ooo(.# — 10) == S7333 ~~ 2489.2:; 

x < 10, EI tan a r = —31 11 1 — 2667.3; + J$6x 2 ; 

x> 10, Eltan a x = — 131055 -f 37333* — 1245* 2 . 

For deflections, 

# < 10, EIv = — 31111^— 1334.x 2 4- 252a: 3 ; 

jp > 10, EIv — — 131055.2: -j- 186673; 2 — 415.x 3 4- *"• 

When x — 15, z; = o ; 

/. .* > 10, £7z/ = —833625 - 131055* 4- 18667a; 2 — 4 I 5^ 3 - 

For maximum deflection, equate slope to zero, and find x. 

We find it at x = 8.41. 
.-. EIv = —24506. 

Hence greatest bending-moment and greatest deflection are 
both in span AB. 

Observe, that, since we have used one foot as our unit of 
measure, all dimensions must be taken in feet, and the value 
of E is also 144 times that ordinarily given. 



582 APPLIED MECHANICS. 

§ 247. Continuous Girder, with both Distributed and 
Concentrated Loads. — In this case we may either calculate 
the bending-moments, slopes, and deflections due to each sepa- 
rately, and then add the results with their proper signs, or we 
may modify the solution that was used for the case of a dis- 
tributed load, so as to extend its applicability to this case. 

Let W represent any one concentrated load, and let x x rep- 
resent the distance of its point of application from the origin. 
Then, in the general formulae deduced for the distributed load, 
make the following changes ; viz., — 

i°. Instead of 

wdx 2 , 
put 



m 



since, as was shown, m represents the sum of the moments of 
the loads, between the section and the support, about the sec- 
tion. 



put 



Jo Jo I Jo Jo ') EP 



x — x 1 )dx i 



EI 



and make the corresponding changes in the values of m lt m_„ 
V lf and V_ 1} leaving n and q just as before ; then use the same 
three-moment equation as before, with these substitutions, i.e., 

M \h. _ * + fci _ 2ril _ u& - M*=i - '*&■- ^ 

2 U* /, /-, 2 Li) h 2 Lf h x . L, 2 



SPECIAL CASE. 583 



SPECIAL CASE, 

when the distributed load is uniformly distributed on each span, 
but may be different on the different spans, and when the girder 
is of uniform section. 

Let w x = weight per unit of length on OA. 
w_j = weight per unit of length on OB. 

Denote by W x any concentrated load on OA at distance x t 
from O. 

Denote by W_ x any concentrated load on OB at distance 
x_! from O, 

Then we shall have 

m. z = — j-i + %W_*{l_ x - x_ x ), 

V - ^L 4 4- s ^( 7 ' ~^) 3 

24^/ r 6-E/ 

and, as before, % 

/, 2 _ /_£ 

*" " 2^7 J n ~ x ~ 7e~i' 

Making these substitutions in the three-moment equation, 
and clearing fractions, we obtain for the case, when the sup- 
ports are all on the same level, 

o = MJ_ t + 2M 2 {h +/_,) + M z k + l(w& + «/_,/_ x 3 ) 

+ j% \ WJIS - (/ x - *)'] W - *) J 

-f JLS J ^-x[/_x 2 - (/_x - *-x) 2 ](/_x - *-,)i- 
'— I 



584 APPLIED MECHANICS. 

CONCENTRATED AND DISTRIBUTED LOADS. 

Example. — Let the girder be of uniform section, of two 
equal spans, each being 10 feet ; let the concentrated loads be 

as shown in the 



c 



4/ ~ & Ji< 4' >l< 3' > figure, the dis- 
tributed load be- 
1000 10()0 . ■ .. 

ing 90 lbs. per 

2000 FlG - 255 - foot. Find the 

value of EI y so that the deflection may nowhere exceed -^\-§ of 
the span. 

Solution. — Use equation (12); and, in deducing value of 
M B , use dimensions in feet ; afterwards use inches. 

Originate M A = M c = o; 

4oM B + 2\ 6 -(iooo + 1000) 

+ yVj 2000(64) (6) + IOOO(5l) (7) + IOOO(9l) ( 3 )5 = O, 

40^ 4- 48000 4- 139800 = o, or M B = —4695 foot-lbs., 

or M B = —56340 inch-lbs., M A = M c = o. 

m^ = 177600, m_ x = 201600, 

7200 «, 60 7200 n_ x 60 

Ul = ~El i 7 L = ei' n ~ x = ~EI ' T^El' 

288000 q r 20 288000 q_i 20 

q * = EI ' 1} = El' q ~* = EI ' U 2 = El' 

175680000 V l 1464000 193536000 V_ z 161 2800 

Vl = EI ' J, = ~~Ef V ~ z = EI ' ~E V = EI 

o 4- 56340 4- 177600 
Shear right side of middle = = 1949.5, 

+ 56340 -f 201600 

Shear left side of middle = = 2149.5 : 

120 y J 

cu . r , , -56340 4- 153600 

Shear left end = = 810.5, 

120 J ' 

— 56340 4- 177600 

Shear risrht end = = 1010.5 : 

& 120 J 

Middle supporting-force = 4099. 






SLOPES. 585 

B ending-Moments in Each Span. 
Span AB, origin at A, 

810.5.2: — 4X 2 
or 

810.5.2: — 4X 2 — 2000 (x — 72). 

Span 2?£ origin at i?, 

-56340 + 1949-5* - 4* 2 > 

— 56340 + 1949-5* — 4* 2 — 1000 (# — ^), 

— 56340 + 1949. $x — 4X 2 — iooo(jc — 36) — iooo(^: — 84). 

To ascertain position of the greatest bending-moment, dif- 
ferentiate each one. 

810.5 ~~ & x = o, x = 101.31 ; 

810.5 — 8x — 2000 =0, X = a minus quantity; 

1949.5 — 8x = o, x = 243.69; 

1949.5 — 83: — 1000 =0, x = 118.69; 

1949.5 — 8x — 1000 — 1000 = 0, x = a minus quantity. 

Hence, in span AB, maximum bending is at the load, and 
its amount is 

(810.5) (72) - 4(72) (72) = 37620. 

Span BC, maximum is at right-hand load, and is 
-56340 + 1949.5(84) - 4(84) (84) - 1000(48) = 31194. 



SLOPES. 

Slope at B, 

rr —5 6 340, (177600) (20) 1464000 165600 

T = -El-^° - 60) -^ + -— - = _^. 

Slope and Deflection in Span AB. 
First part, 

tana = -gj J 405.25^ - -* 3 | + tana , 
a being slope at A. 



586 APPLIED MECHANICS. 



Second part, 

1 ( 4 ) 

tana = -pjK 405.25^ — -x 3 — 10003 2 -f 1440003: > -f- c. 

When x = 72, a is the same in both cases ; 

/. ^-|iooo(72) 2 — (144000) (72) J -f tana — c = o 
EI 

5184000 
/. c = tana — — jr- 

165600 
When x = 120, the second value of tana becomes —^7 — ; 

EI 

""' £I\ ^ 4 °5- 2 5) (l20) 2 --(l20)3-(lOOO)(l20) 2 + I 44 000(l20) | 

5184OOO 16560O 



-+- tan a Q — 



EI ~ EI 
1062000 



tana = -^-f— 641 1600 + 5184000 + 165600J = — 

EI EI 

6246000 

Hence slope in first part (between A and the load), 

tana = -=rf\ —1062000 + 405. 253^ — -^ 3 |* 

Second part (between B and the load), 

tana = —\ —6246000 -h 1440003: — 594-75^ 2 — \ x \ 

Deflection. 
First part, 

v = — \ — 10620003 + 135.083: 3 I. 

EI\ 3 ) 

Second part, 

v = — i —62460003: + 720003 2 — 198.2533 — -x*l + 
^/( 3 ) 



SLOPES. 587 



When x = 1 20, z; = o ; 






1 ( (120) 4 ) 
-^ J (6246000) (120) - (72000) (i2o) 2 + (198.25) (I20)3H — \ 

120/ , v I244l6000 

= — (1036800) = — 22 . 

EI y 6 J EI 

Point of greatest deflection is found by putting slope equal 
zero. Moreover, it is plain that the greatest deflection is in the 
first, and not the second, part. 

Hence equation is 

f* 3 — 405.25.x 2 4- 1062000 = o 
/. x= 56-77; 

and, substituting this in the expression for the deflection, we 
obtain 

39037720 



Vq = 



EI 



I2 o 39037720 . . 

Hence, putting = =^= — , we obtain 

r & 400 hi 

EI — 130125733. 

If E = 1400000, / = 92.9 ; therefore, if b = 3 inches, h = 
7 inches. 

Slope and Deflection in Span BC. 
Portion nearest &, 

tana = — J 165600 — 56340* -f 974.8** * 3 >- 

When x — 36 inches, we obtain 

t 661507 
tana = -^(165600 — 2028240 + 1263341 — 62208) = -=j — 

Middle portion, 

1 ( 4 ) 

tana = -gjl -20340* + 474-75* 2 — ~ xZ \ + *- 



588 APPLIED MECHANICS. 

661507 

When x = 36 inches, then tan a = — j=j — ; 

— 661507 = —732240 4- 615276 — 62208 4- EIc 
482335 
" ■ EI 

.-. tana = 2=y j -482335 - 20340* -f 474-75^ 2 — ~* 3 j* 

When x — 84 inches, 

tana= ^(-482335 - 1708560 + 3349836 - 790272) 

368669 
~ + EI 
Portion nearest C, 

i — -pj\ 63660* — 25** x 3 > + c. 

368669 
When .r = 84 inches, then tana = — wr~ > 

.*. 368669 = 5347440 — 176400 — 790272 4- EIc 

4012099 
•'• ' = £T~ 

.*. tana = -p-jl —4012099 4- 63660* — 25** * 3 >• 

When x = 120 inches, 

1 963101 

tana = -^(—4012099 4- 7639200 — 360000 — 2304000) = — p. • 

Deflection. 

Portion nearest B, 

v = — — \ 165600* — 281 70* 2 4- 324-9* 3 * 4 >. 

EI { 3 ) 

When *- = 36 inches, 

v = -^- (165600 - 1014120 4- 421070 - 15552) (36) 

(443002)36 _ 15948072 
~~ EI ' ~ -5/ 



SLOPES. 589 

Middle portion, 

v = < —482335.x — 10170* 2 4- 158.25*3 * 4 1 + c. 

15948072 
When x = 36 inches, then v = pj — ; 

•'• -1594807 2 = (-482335 - 366120 + 205092 - 15552) + EIc, 

15289157 






EI 



.-. v = ^j\ -15289157 -482335* - 10170* 2 + 158.25*3 - -*H. 

Greatest deflection occurs in the middle portion, and the 
point is given by the equation. 

o = -482335 - 20340* + 474.75** - |*3 = o; 

.♦. * = 71.4. 

Greatest deflection in span BC> 



Fig. 256. 
V = 

-^(-15289157-34438719-51846253 + 57602105-8662899) 

52634923 

~ EI 

120 52634023 

Hence, putting = p/-, we obtain 

400 EI 

EI= 175449743; 

therefore, if E = 1400000, we have 

/= 125.3. 

If b = 3 inches, h = 8 inches. 



590 APPLIED MECHANICS. 



EXAMPLES OF CONTINUOUS GIRDERS. 

i°. Let I = uniform moment of inertia of girder. 

w = load per unit of length uniformly distributed. 
Find expressions for 

1, the bending-moment over each support, 

2, the supporting-forces, 

3, the greatest bending-moment, 

4, the slopes at the supports, 

5, the greatest deflection, 

in each of the following cases : — 

(a) Two equal spans, length /. 

(b) Three equal spans, length /. 

(c) Four equal spans, length /. 

(d) Two spans, lengths 4 and 4 respectively. 

(e) Three spans, lengths /„ / 2 , and / 3 respectively. 
(/) Four spans, lengths /,, l 2 , l v and / 4 respectively. 

(g) Two equal spans ; loads per unit of length on each span, 
w x and w 2 respectively. 

(h) Three equal spans ; loads per unit of length on each span, 
w iy w 2 , and w 3 respectively. 

2°. Do the same in the case where each span is loaded with 
a centre load W, and has no distributed load. 

3°. Find greatest bending-moment and greatest deflection 
for a continuous girder of two spans, uniformly loaded on these 
two spans with load w per unit of length, and which overhang 
the outer supports ; the overhanging parts having lengths l_ Q 
and 4 respectively, and the same distributed load per unit of 
length on the overhanging parts. 



EQUILIBRIUM CURVES. — ARCHES AND DOMES. 59I 



CHAPTER IX. 



EQUILIBRIUM CURVES. — ARCHES AND DOMES. 



§ 248. Loaded Chain or Cord. — It has been already shown 
(§ 126), when the form of a polygonal frame is given, that the 
loads must be adapted, in direction and magnitude, to that form, 
or else the frame will not be stable. The same is true of a 
loaded chain or cord, which would be realized if the frame were 
inverted. 

If a set of loads be applied which are not consistent with 
the equilibrium of the frame under that form, it will change its 
shape until it assumes a form which is in equilibrium under the 
applied loads. 

As to the manner of finding (when a sufficient number of 
conditions are given) the stresses («) <j>) 

in the different members, etc., this 
was sufficiently explained under the 
head of "Frames," and will not be 
repeated here, as the figures speak 
for themselves. 

In Fig. 257 the polygon fedcbaf 
is the force polygon, while the equilibrium polygon is 123456, 
an open polygon. A straight line joining e and a would repre- 
sent the resultant of the loads. 




Fig. 257. 



592 



APPLIED MECHANICS. 



CHAIN WITH VERTICAL LOADS. 

If all the loads are vertical, the broken line edcba becomes 

a straight line and vertical, as 
shown in Fig. 258$. Wheneverthe 
loads are concentrated at single 
points, as 2, 3, 4, 5, the form of 
the chain is polygonal ; and when 
the load is distributed, it becomes 
a curve, as shown in Fig. 259. 




Fig. 258. 




CURVED CHAIN WITH A VERTICAL DISTRIBUTED LOAD. 

Given the form of the chain AOE supported at A and E, 
and the total load 
upon it (be, Fig. 
259^), to find the 
distribution of the 
load graphically. 
First lay off be to 
scale, to represent 
the total load : this 
is balanced by the two supporting-forces at A and E respec- 
tively, as shown in the figure. Hence draw ca parallel to the 
tangent at E, and ba parallel to that at A, and we have the 
force polygon abca ; the equilibrium curve being the chain' AOE 
itself. Moreover, if the lowest point of the chain be O, then 
the load must be so distributed that the portion between O and 
A shall be balanced by the tension at O and that at A, and 
hence that its resultant shall pass through the intersection of 
the tangents at O and A. Its amount will be found by drawing 
from a a horizontal line ; and then we shall have ao as the ten- 
sion at 0, ab as the tension at A, and bo as the load between A 
and 0. Hence the load between E and O will be oc. 






LOADED CHAIN OR CORD. 593 

Moreover, the load between O and any point, as B, will be 
balanced by the tension at O, and the tension at B, and hence 
will be od, where ad is drawn parallel to the tangent BD, so 
that the load between b and c will be dc ; and in this way we see 
that we can find the tension at any point of the chain by simply 
drawing a line from a, parallel to the tangent at that point, till 
it meets the load-line be. 

It is to be observed, that, if the tension at any point of the 
chain be resolved into horizontal and vertical components, the 
horizontal component will, when the loads are all vertical, be a 
constant, and the vertical component will be equal to the por- 
tion of the load between the lowest point and the point in 
question. 

If we assume our origin at O, axis of x horizontal and axis 
of y vertical, and let the co-ordinates of B be x and y, and if w 
be the intensity of the load at the point (x, y), we shall have, for 
the load od between O and B, 

P = fwdx; 



and, since the angle oad = angle BDC> we shall have 

dy_ _ BC^ _ od _ P_ 
ax ~" Z1C ~ ' o~a~ ' ~H 



By differentiation, we shall have 



or 



d 2 y _ 


\dx) 


dx 2 


H 


d 2 y _ 
dx 2 ~ 


w 



(I) 



and this is the equation for all vertically loaded cords. 



594 



APPLIED MECHANICS. 



From it we can find the form of the cord to suit a given 
distribution of the load. 

§ 249. Chain with the Load Uni- 
formly Distributed Horizontally. — 
In this case w is a constant ; and if 
we assume our origin at the lowest 
point of the chain, and use the same 




Fig. 260. 



notation as before, we shall have 

d 2 y _ 
dx 2 



w 



Hence, integrating, and observing, that, when x = o, — = o, 
we have 



dy_ 
dx 



wx 



and by another integration, observing, that, when x = 0, y = 0, 
we obtain 

y = x 2 . 

y 2H 

This is the equation of a parabola ; hence a chain so loaded 
assumes a parabolic form. 

Example I. — Given the heights of the piers for support- 
ing a chain so loaded, above the lowest point of the chain, as 
8 and 18 feet respectively, the span being 100 feet, to find the 
distance of the lowest point from the foot of each pier, and 
the equation of the curve assumed by the chain. 

Solution. — If (with the lowest point of the chain as origin) 
we call (x If j/ t ) the co-ordinates of the top of the first pier, and 
(x 2 , f 2 ) those of the top of the second pier, we shall have, since 
j/j =: 18 and y 2 = 8, and since we must have 



2& 



CHAIN WITH UNIFORM HORIZONTAL LOAD. 595 





18 = — x T 2 and 8 = —r-x 2 2 
2H 2H 


x 2 ▼ 


/^ _ 3 . ._ 3 ; L ._ s . 

' — . . x z X 2 . « «#i -+" ^2 — -#2 3 

02 2 2 


but 




Xj, ~\~ x 2 — 


IOO .*. |-^ 2 = IOO .*. X 2 =40, X- = 60. 



Hence, since 18 = — -(60) 
2// 



w 



2H 360O 200' 

therefore equation of the curve is 

y = y*o* 2 - 

Example II. — Given the load on the above chain as 4000 
lbs. per foot of horizontal length, to find the tension at the low- 
est point, also that at each end. 

Solution, 

w 1 

-— = — , w = 4000, 

2H 200 

.*. 2H — 800000 ,\ H = 400000 lbs. 

Moreover, load between lowest point and highest pier = 
60 X 4000 = 240000 lbs. 

Therefore tension at highest pier = 



y/(24000o) 2 -J- (400000) 2 = iooooy/(24) 2 -f- (40) 2 

= iooooy/2176 = 466480 lbs. 
Tension at lowest pier = 



y/(i6oooo) 2 -f- (400000) 2 = 10000^256 + 1600 

= 10000^1856 = 437919 lbs. 



596 APPLIED MECHANICS. 

Example III. — Given the span of the chain as 20 feet, and 
its length as 25 feet, the two points of support being on the 
same level, to find the position of the lowest point. 

§ 250. Catenary. — The catenary is the form of the curve 
of a chain, which, being of uniform section, is loaded with its 
own weight only, i.e., with a load uniformly distributed along 
the length of the chain. 

To deduce the equation of the catenary : if we assume the 
origin, as before, at the lowest point of the curve, we shall 
have still the general equation 

d 2 y _ w 
~dx~ 2 ~ H' 

but w in this case is not constant. 

If we let Wj = the load per unit of length of chain, we 
shall have 



ds 

w = w T — = w 
dx 



■€m-. 



hence 



Or, if we let 



a constant, 



d 2 y w x ds 

dx 2 ~~ ~H~dx 

a/,_ 1 
H~ m 

d 2 y _ 1 ds 
dx 2 m dx 1 



(1) 



which is the differential equation of the catenary ; and we only 
need to integrate it to obtain the equation itself. 
To do this, we have 

d 2 y 
rf2y dx 2 1 

dx 2 



m)i \dx) 



■' vHS) 



THE CATENARY. 



197 



therefore, integrating, and observing, that, when x = o, -y- = o, 

ax 

we shall have 



-(SH--"-£+(t 



(2) 



/. -J. = -Je»* — e m ) /. y = — ( e m + * w ) -+■ c 

dx 2\ / 2\ / 

But, when ;tr = o, y = o .'. c == —m; hence the equation is 

(3) 



m/ x_ _x\ 
y = —\e m + e ** J 



2 \ 



and this is the equation of the catenary 
when the origin is taken at O, the low- 
est point of the chain. 

If it be transferred to O s , where OO t 
== m i the equation becomes (by putting 
for y t y — m) 

ml * _*\ 
y — — ( e m — e m \. 





Y 


E 


v^ 


^ 


o 


C 


— X, 



Fig. 261. 



(4) 



This is the most common form of the equation to the cate- 
nary, the origin being taken, at a distance below the lowest 

TT 

point of the curve equal to m =■ — , the horizontal tension 
divided by the weight per unit of length of chain. 



598 APPLIED MECHANICS. 



To find x in terms of y, we have 
«S + JL = 2 

— tn 

e m 

■zx 2V ■*- ** 2V x 

.*. e m -f" 1 = —e m s. em — _£ ^?» = — j, 



Solving, we have 



X_ y 

e ™ — t- ± 

m 



Jf-i ••• ^=logJZ ±V /^_ I ) 

▼ m 2 m \m V m 2 ) 

.-. * = «log r U±t/^_ I l. (5) 

To find the length of the rope : from the equation 

j = — ( em -\- e m ] 

we obtain 

dy 1/* _*\ 

^r 2\ / 



^vSS^^+K^-^^ 



dx y ^dx 



/I / 2£ _H£\ I / £ :£'\ 

= V/-( *? + 2 + ^ m \ = -l e m + e ~ m\ 

.'. -^- = -(e^ + e~m) (6) 

dx 2\ ' 

,\ s = - / [e m + e~m)dx ——{em — e~mj. (7) 

To find the area OO.AB, we have 

X-* J7Z /•■*/ ■* £\ #2 2 / * _^\ /ox 

ydx — — I (<?™ + £ < )a& = — [em — * «l. (8) 






But 



TRANSFORMED CATENARY. 599 



arc OB — —\e m — e ™) = s; 



hence area OO x AB = ms. 

This shows, that, if the load should be distributed in such a 
way as to be like a uniformly thick sheet of metal, having for 
one side the catenary and for the other the straight line O x A, 
the equilibrium curve would be a catenary. 

It may be convenient to have the development of e m and 

e~™\ hence they will be written here : — 

-f- -4r~ + 7^7- + etc., 



e ™ = I + - + 

m 


x 2 - 
m 2 \2 


e~™ = 1 - — 4- 
m 


X 2 

m 2 \2 



fn*\$ m 4 ^ 



x* , x 4 , , 

H 77- + etC * 



m 2 \2 m^ w 4 |4 



Example I: — Given a rope 90 feet long, spanning a hori- 
zontal distance of 75 feet; find the equation of the catenary, 
the sag of the rope, and the inclination of the rope at each 
support, supposing these to be on the same level. 

§251. Transformed Catenary. — We have just seen that 
the catenary is the form of chain suited to a load which may 
be represented by a uniformly thick sheet of metal, with a hori- 
zontal extrados, provided the distance 00 I is equal to m, a defi- 
nite quantity. A more genera] case, however, would be that of 
a chain loaded with a load which might be represented by a 
uniformly thick sheet of metal, where the length 00, is any 
given quantity whatever. A chain so loaded is called a trans- 
formed catenary, and the catenary itself becomes a particular 
case of the transformed catenary. 

We may deduce its equation as follows : — 



6oo 



APPLIED MECHANICS. 



Let the chain be represented by ACB> and let it be so 
loaded that the load on CD is repre- 
sented by .w times area OCDE> so that 
w = weight per unit of area ; then we 
shall have, for this load, 




= w I ydx. 



Hence, from what we have already seen, 



dy_ 
dx 



_ ^ _ ™ c 



ydx 



d 2 y w 
dx~ 2 = Ti- 



dy d 2 y w dy 

dx dx 2 H J dx 



Hence, integrating, we have 



m- 



w 
H 



f + c. 



dy 
But, when -=- = o, y = a ; 



w . 



(i)' =>-••>■ 



Or, if we write, for brevity, — == m 2 , we have 

w 



M V _ y 2 ~ <? 
\dx) ' m 2 



• — = — sjy 2 — a 2 
dx m 



dy 



dx 



Sly 2 = a 2 m 



\ \og(y + >Jy 2 - a 2 ) = - + c 



LINEAR ARCH. 601 



But, when x = o, y = a; 
:. log (a) = c 



(y + Vr - « 2 ) i * 

7/z 






y + Vjv 2 — # 2 * zx x_ 

= e m .*. jy 2 — # 2 = #V W — 2aye m -\- y 2 



2V x x 

— = ^w -f- £ ** 







which is the equation of the transformed catenary. This 
becomes the catenary itself whenever a = m. 

Example. — Given a chain loaded so that the load on OD is 
proportional to the area OEDC. Let OC = 5 feet, BF— 8 feet, 
OF = 4 feet ; weight per unit of area == 8o lbs. Find the 
equation of the transformed catenary, also the tension at C and 
that at B. 

§ 252. Linear Arch. — In all the preceding cases, the chain 
or cord is called upon to resist a tensile stress arising from a 
load that is hung upon it. If, now, the cord be inverted, we 
have the proper equilibrium curve for a load placed upon it, dis- 
tributed in the same manner as before ; only in this latter case 
the cord would be subjected to direct compression throughout 
its whole extent. The equilibrium curve is, then, sometimes 
called a linear arch. The general equation of the equilibrium 
curve remains just as before, 

d 2 y w 

the axes being so chosen that OX is horizontal and Y verti- 
cal. 

Thus, if it were required to find the form of the equilibrium 
curve or linear arch, with the upper boundary of the loading 
horizontal, we should obtain a transformed catenary. 



602 APPLIED MECHANICS. 

§ 253. Arches. — In the case of arches composed of a series 
of blocks, as in stone or brick arches, the mathematical treat- 
ment generally used for determining the proper form and 
proportions of the arch has been quite different from that used 
for the determination of the proper form and proportions of 
the iron arch, whether made in one piece, or two pieces hinged 
together, or of a lattice. 

In the case of the iron arch, the treatment involves neces- 
sarily a determination of the stresses acting in all its parts, and 
an adaptation of its form and dimensions to the load, so that at 
no point shall the stress exceed the working-strength of the 
material. 

In the case of the stone arch, it is still a question under 
discussion whether it would not be best to adopt the same 
method, although it would lead to a great deal of complexity, on 
account of the joints. 

Nevertheless, the question usually raised is one merely of 
stability ; i.e., as to the proper form and dimensions to pre- 
vent, not the crushing of the stone, though this must also be 
taken into account if there is any danger of exceeding it, but 
more especially the overturning about some of the joints. 

The question of the stability of the stone arch may present 
itself in either of the two following ways : — 

i°. Given the arch and its load, to determine whether it is 
stable or not. 

2°. Given the distribution of the load, to determine the 
suitable equilibrium curve, and hence the form of arch, suited 
to bear the given load with the greatest economy of material. 

§ 254. Modes of giving Way of Stone Arches. — An arch 
may yield, (i°) by the crushing of the stone, (2 ) by sliding of 
the joints, (3 ) by overturning around a joint. The following 
figures show the modes of giving way of an arch by the last 
two methods. The first two show the dislocation of the arch 
by the slipping of the voussoirs. In the former case the 



FRICTION. 



603 



haunches of the arch slide out, and the crown slips down ; in 
the other case the reverse happens. The second two figures 
show the two methods by which an arch may give way by 
rotation of the voussoirs around the joints. 





Fig. 263. 



Fig. 264. 





Fig. 265. 



Fig. 266. 




Before proceeding farther with the problem of the arch, two 
or three matters of a more general nature will be treated, 
which will be necessary in its discussion. 

§ 255. Friction. — Let AB be a plane inclined to the hori- 
zon at an angle 0. Let D be a body resting 
on the plane, of weight DG = W. Resolve 
W into two components, DE and DF respec- 
tively, perpendicular and parallel to the 
plane. The component DE = WcosO is 
entirely neutralized by the re-action of the plane ; while DF 
= Ws'm 9, on the other hand, is the only force tending to make 
the body slide down the plane. It is an experimental fact, that 
when the angle is less than a certain angle </>, called the 
angle of repose, the body Soes not slide ; when 6 = <£, the body 
is just on the point of sliding ; and when is greater than <£, 
the body slides down the plane with an accelerated motion, 
showing that in this case an unbalanced force is acting. This 



Fig. 267. 




9 



604 APPLIED MECHANICS. 

angle <f> depends upon the nature of the material of the plane 
and of the body, and on the nature of the surfaces. Hence, 
in the first and second cases, the friction actually developed 
by the normal pressure DE just balances the tangential com- 
ponent DF; whereas, in the third case, when the angle of 
inclination of the plane to the horizon is greater than <£, the 
tangential component DF is only partially balanced by the 
friction. 

Let ab be the plane when inclined to the horizon at an 
angle <f>. The body is then just on the 
point of sliding, hence the component 
df — Wsin<£ is just equal to the fric- 
tion developed between the two surfaces. 
fig. 268. Moreover, if we represent by N the 

normal pressure de = Wcos <f> on the plane, we shall have 

df — iVtan $. 

Now, it is an experimental fact, that the friction developed 
between two given surfaces depends only on the normal press- 
ure, i.e., that the friction bears a constant ratio to the normal 
pressure; and since, in this case, the friction just balances the 
tangential component df = iVtan <f>, the friction due to the 
normal pressure iV is 

jVtan <f>. 

Now, it makes no difference what be the position of the 
plane surface : if a normal pressure N be exerted, the friction 
that is capable of being* exerted to resist any force F tangential 
to the plane, tending to make the bodies slide upon each other, 
is N tan <f> ; and if the force F is greater than ^Vtan </>, the bodies 
will slide, but if Fis less than TV tan <£, they will not slide. The 
quantity tan <£ is called the co-efficient of friction, and will be 
denoted by f. 




STABILITY OF BUTTRESS ABOUT A PLANE JOINT. 605 

From the preceding it is evident, that, if the resultant press- 
ure on the body makes with the normal to the plane an angle 
less than the angle of repose, the sliding will not take place ; 
whereas, if the resultant force makes with the normal to that 
plane an angle greater than the angle of repose, the body will 
slide. 

§ 256. Stability of Position. — To determine under what 
conditions the stability of the block 
DGHF is secure against turning around 
the edge D: if the resultant of the 
weight of the block and the pressure 
thereon pass outside the edge D> as OR„ 
then the block will overturn ; the mo- 
ment of the couple tending to overturn it 
being OR, X DE. If, on the other 
hand, it pass within the edge, as OR 2 , the block will not over- 
turn, since the force has a tendency to turn it the opposite 
way around D. Hence, in order that a block may not overturn 
around an edge at a plane joint, the resultant pressure must 
cut the joint within the joint itself. 

In any structure composed of blocks united at plane joints, 
we must have both stability of position ond stability of friction 
at each joint, in order that the structure may not give way. 

§257. Stability of a Buttress about a Plane Joint. — 
Let DCEF be a vertical section of a buttress, against which 
a strut rib or piece of framework abuts, exerting a thrust 
P = ZG = OR. In order that the buttress may not give 
way, it must fulfil the conditions of stability at each joint. Let 
AB be a joint. Should several pressures act against the but- 
tress, the force P in the line ZO may be taken to represent 
the resultant of all the thrusts which act on the buttress above 
the joint AB. Let G be the centre of gravity of the part 
ABEF, and let W — OL be the weight of that part of the 
buttress. Let be the point of intersection of the line of 



6o6 



APPLIED MECHANICS. 




direction of the thrust, and of the weight W. Draw the paral- 
lelogram ORNL. Then will ON be the resultant pressure on 
the joint AB : and the conditions of stability require that the 
resultant pressure should cut the joint 
AB at some point between A and B, and 
that its line of direction should make 
with the normal to AB an angle less 
than the angle of repose, <f> ; and, in 
order that the buttress may not give way, 
these conditions must be fulfilled at each 
and every joint. 

Another way of expressing this con- 
dition is as follows : The force tending 
to overturn the upper part of the but- 
tress around A is the force F = OR ; 
and its moment around A is F(Ar) = Fp 
if we let Ar = p, whereas the moment of the weight which 
resists this is W{AS) = Wq if we let AS = q. Now, when 
ON passes through A, we have Fp = Wq ; when ON passes 
inside of A, we have Wq > Fp ; when ON passes outside of A, 
we have Wq < Fp. Hence the conditions of stability require 
that 

Wq>Fp or Fp < Wq. 



Example. — Given a rectangular buttress 
8 feet high, I foot wide, and 4 feet thick; the 
weight of the material being 100 lbs. per 
cubie foot, the buttress being composed of 8 
rectangular blocks 1 X 4 X 1 foot. On this 
buttress is a load of 500 lbs., whose weight 
acts through K, where OK = 3 feet. Find 
the greatest horizontal pressure P that can be applied along 
the line OK, consistent with stability, against overturning 
around each of the edges a, b, c, d, <?,/, g, h. 




Fig. 271. 



LINE OF RESISTANCE IN A STONE ARCH. 6o? 

Solution. — The weight of each block will be 400 lbs. 
Hence we shall have the following equations : — 



1500 4- 400 x 2 
Stability about a, max P — ■ = 2300. 



= i55o. 

= 1300. 

-= 1 1 75. 
= 1 100. 



u a 











I 


it 




I5OO 


-f 800 X 2 










2 


11 




I500 


+ 


I200 X rf 2 










3 


tt 




I5OO 


+ 


l600 X 2 










4 


tt 




I5OO 


+ 


2000 X 2 










5 


(i 




I500 


+ 


2400 X 2 










6 


te 




1500 


+ 


2800 X 2 










7 


a 




I500 


+ 


3200 X 2 



" /, " = r = io 5°- 



= 1014. 



h, " = - — i = 987. 



The least of these being 987 lbs., it follows that the great- 
est pressure consistent with stability is 987 lbs. 

§ 258. Line of Resistance in a Stone Arch. — In order 
to solve any problem involving the stability of a stone arch, it 
is necessary that the student should be able to draw a line of 
resistance. To make plain the meaning of the term, the follow- 
ing solution of an example is given. The method of drawing 
the line of resistance employed in this solution is given purely 
for purposes of illustration, and is not recommended for use in 
practice, as a suitable method will be given later. 



6o8 



APPLIED MECHANICS. 



Example. — Given three blocks of stone of the form shown 

in the figure (Fig. 272), their 
common thickness (perpendic- 
ular to the plane of the paper) 
being such that the weight 
per square inch of area (in the 
plane of the paper) is just one 
pound. 

Given AC = 13 inches, 
BC = 8 inches. Suppose 
these three blocks to be kept 
from overturning by a hori- 
zontal force applied at the 
middle of DE. Find the least 
value of this horizontal force consistent with stability about 
the inner joints, also its greatest value consistent with stabil- 
ity about the outer joints. 
Solution. 

BK = 16 sin 15 = 4.14112. 
AH = 26sini5° = 6.72932. 

rJ? 2( (, 3 )3- (8)3 ) _ 

3l('3) 2 - (8)M ~ 















E 


V 


~G> 


L 


M 


/ ? 


/ 




G 2 ---V 


P' 

D, 
P, 
G, 










f H, 




/ / K 3" 


Y/ 




._-J^____J Kl 


/ P v 

/ / 
/ / 


/ 








P. 1 
C 


A 


B 











Fig. 272. 



TO. 7. 



Altitude of each trapezoid = 5 cos 15 
Area of each trapezoid = x ^> sin 30 
Weight of each stone 

GG 2 = 8sin30° — 10.7 cos 15 sin 15 
KK 2 = 8COS30 — 10.7 cos 15 sin 15 
KK Z = 10.7 cos 15 cos 45 — 8COS30 
BN 2 = 8 — 10.7 cos 15 sin 15 
BN Z = 8 — 10.7 cos 15 cos 45 
BN A = 10.7 cos 15 sin 15 
HH 2 = 13COS30 — 10.7 cos 15 sin 15 



HH % 



13 cos 30 — 10.7 cos 15 cos 45 = 



4.8296. 
26.25 sq. in. 
26.25 lbs. 

1-325 
4-253 
0.380 

5-325 
0.692 
2.675 
8.583 
3-95° 



LINE OF RESISTANCE IN A STONE ARCH. 



609 



AN 2 as 13. — 10.7 cos 15° sin 15° 


= io-3 2 5- 


AN 3 = 13 — 10.7 cos 15 cos 45 


= 5.692. 


AN 4 = 13 — 10.7 cos 2 15 


= 3- OI 7- 


GjM = 10.5 — 8COS30 


= 3-572. 


X,M = 10.5 - 8sin30° 


= 6.500. 


Ol/ = 10^5 


= 10. 500. 


H^M ~ 10.5 — i3sin30° 


= 4.000. 



Let us represent the thrust at M by 71 Then, to find what 
is the thrust required to produce equilibrium about G, we take 
moments about G, and likewise for the other joints. We may 
proceed as follows : — 

INNER JOINTS. 

Stability about £, 
T(GM) = (26.25) (GG 2 ) 

7(3.572) = (26.25) (1.325) .-. T= 9.74. 

Stability about K, 

T(KJf) = (26.25) (KK 2 - KK 2 ) 

7(6.500) = (26.25) (4.253 - 0.380) ,\ 7 7 = 15.49. 

Stability about B, 

T(CM) = (26.25) (BN 2 + BN Z - BN 4 ) .-. 7= 8. 3 6. 



or 



or 



OUTER JOINTS. 

Stability about H } 

T(H,M) = (26.25) (HH 2 -f- J7iy 3 ) ... r= 82.25. 

Stability about ^, 

T(CM) = (26.25) (AN 2 + ^ 3 + AN,) .-. r= 47-59 

It is plain, therefore, that, in order to have equilibrium, the 



6io 



APPLIED MECHANICS. 



thrust at M must be between 15.49 lbs. and 47.59 lbs. : for, 
if it is less than 15.49 lbs., the arch will turn about an inner 

joint ; and if it is greater than 
47.59 lbs., it will turn around 

an outer joint. 

If, now, we draw through 
M a horizontal line to meet 
the vertical drawn through the 
centre of gravity of the first 
stone, and lay off ajB = 15.49, 
and ay = 26.25, then will the 
resultant of this thrust a/3 and 
the weight of the first stone ay 
be aS; this being the resultant 
•pressure on the joint FG, its 
point of application being c. 
Next, prolong this line aS to 
meet the vertical through the 
centre of gravity of the second 
stone, and combine aS with the 
weight of the second stone, 
thus obtaining, as resultant 
pressure on the joint A77, the 
force £77, whose point of appli- 
cation is at K. Compounding, 
now, £77 with the weight of the 
third stone, we obtain, as final 
resultant pressure on AB, the 
force V applied at p. Now, 





1 1 


p M 


h 


A 


1 
1 


. /' / 


' 



Fig. 273. 



joining McKp by a broken line, we have the Line of Resistance 
corresponding to the thrust 15.49, or the minimum horizontal 
thrust at M. If, now, we construct a line of resistance with 
47.59 lbs., we obtain the line M<r<pA, corresponding to maximum 
horizontal thrust at M. 



SYMMETRICAL DISTRIBUTION OF THE LOAD. 6ll 

If the arch is in equilibrium, and if the horizontal thrust is 
applied at M, it is plain that the actual thrust would either be 
one of these two or else somewhere between these two, and 
hence, that, if the requisite thrust is furnished at M to keep 
the arch in equilibrium, the true line of resistance cannot lie 
outside of these two ; viz., the line corresponding to maximum 
and that corresponding to minimum horizontal thrust at M. 

If the separate stones supported loads, it would be neces- 
sary to take into account these loads, in addition to the weights 
of the stones, in determining the horizontal thrust, and drawing 
the lines of resistance. 

§ 259. Arches with Symmetrical Distribution of the 
Load. — Before considering the conditions of 
stability of an arch, we shall proceed to some 
propositions about lines of resistance corre- 
sponding to maximum and minimum horizon- 
tal thrust. If, in an arch, we draw a line of 
resistance AB through the point A of the 
crown, and then, by changing the horizontal ^bb« 
thrust, we change the line of resistance con- FlG ' 2?4 ' 

tinuously till it touches the extrados of the arch at C, we 
shall evidently have, in the line AC'B r , a line of resistance 
which has the greatest horizontal thrust of any line that passes 
through A, and lies wholly within the arch-ring. If, on the 
other hand, we decrease gradually the horizontal thrust until 
the line touches the intrados at D\ then we have in this line 
the line of minimum horizontal thrust that passes through A. 
By lowering the point A, however, and keeping the point £7 the 
same, we should obtain new lines of resistance with greater 
and greater horizontal thrust ; the greatest being attained when 
the line comes to have one point in common with the intrados. 
Hence a line of maximum horizontal thrust will have one point 
in common with the extrados and one point in common with 
the intrados, the latter being above the former. 




6 12 APPLIED MECHANICS. 

On the other hand, by retaining the point D' the same, and 
raising the point A, we should decrease the horizontal thrust, 
and thus obtain lines of resistance with less and less horizontal 
thrust ; the least being attained when the line of resistance 
comes to have a point in common with the extrados. Hence 
the minimum line of resistance has a point in common with the 
extrados and one in common with the intrados, the latter beine 
below the former. 

These cases are exhibited in the following figures : — 

minimum 
maximum 





Fie 275. Fig. 276. 

§ 260. Conditions of Stability. — The question of the sta- 
bility of an arch must depend upon the position of its true 
line of resistance. If this true line of resistance lies within 
the arch-ring, the arch will be stable provided the material 
of which it is made is incompressible. If this is not the case, 
the stability of the arch will depend upon how near the true 
line of resistance' approaches the edge of the joints ; for the 
nearer it approaches the edge of a joint, the greater the inten- 
sity of the compressive stress at that joint, and the greater the 
danger that the crushing-strength of the stpne will be exceeded 
at that joint. Thus, if the true line of resistance cuts any 
given joint at its centre of gravity, the stress upon that joint 
will be uniformly distributed over the joint. If, however, it 
cuts the joint to one side of its centre of gravity, the intensity 
of the stress will be greater on that side than on the opposite 
side ; and, if it is carried far enough to one side, we may even 
have tension on the other side. 



CRITERION OF SAFETY FOR AN ARCH. 



613 



§ 261. Criterion of Safety for an Arch. — There are two 
criteria of safety for an arch, that have been used : — 

i°. That the line of resistance should cut each joint within 
such limits that the crushing-strength of the stone should not 
be exceeded by the stress on any part of the joint. 

2°. That, inasmuch as the joint is not suited to bear tension 
at any point, there should be no tension to resist. 

The distribution of the stress is assumed to be uniformly 
varying from some line in the plane of the joint. The three 
following figures will, on this supposition, represent the three 
cases : — 

i°. When the stress is wholly compression. 

2°. When the stress becomes zero at the edge B. 

3 . When the stress becomes negative or tensile at B. 



4R1 






In all three figures, AB represents the joint which is as- 
sumed to be rectangular in section, AD represents the intensity 
of the stress at A, and BE that of the stress at B ; while R repre- 
sents the point of application of the resultant stress, RR X rep- 
resenting that resultant. 

Proposition. — If the stress on a rectangular joint vary 
uniformly from a line parallel to one edge, the condition that 
there shall be no tension on any part requires that the result- 
ant of the compressive stress shall be limited to the middle 
third of the joint. 

Proof. — Let AB (Fig. 278) represent the projection of 
the joint on the plane of the paper. It is assumed that the 



6 14 APPLIED MECHANICS. 

stress is uniformly varying ; and, if there is to be no tension 
anywhere, the intensity at one edge must not have a value less 
than zero, hence at the limiting case the value must be zero ; 
hence this limiting case is correctly represented by the figure, 
and the resultant of the compression will be for this case at the 
centre of stress. Thus, if AD represent the greatest intensity 
of the stress, then we shall have, if B be the origin and BA the 
axis of x, if the axis of y be perpendicular to AB at B, and if 
we let a — intensity of stress at a unit's distance from B y that 
RR X = affxdxdy, and (BR) (RR t ) = affx 2 dxdy; 

bk*_ 

ffx 2 dxdy 3 
•'• BR = Jfxdxdy = M> = **» 

2 

if b = breadth, and h = BA "= height of rectangle. 

Hence, if the resultant of the compression be nearer A than 
R, there will be tension at B ; and, on the other hand, if it be 
nearer B than \h> there will be tension at A. Hence follows 
the proposition as already stated. 

While the above is probably the condition most generally 
used to determine the stability of an arch, at the same time, if 
there is any danger that the intensity of the stress at any part 
of any joint may exceed the working compressive strength of 
the stone, this ought to be examined, and hence a formula by 
which it may be done will be deduced. 

Let AB (Fig. 279) be the joint, and let, as before, b be its 
breadth, and h — AB = depth ; then, suppose the pressure to 
be uniformly varying, DA ==/== the working-strength per unit 
of area == greatest allowable intensity of compression ; then the 
entire stress on the joint will be represented by the triangle 
ACD, for the joint is incapable of resisting tension. 
Hence 

AR = \AC :. AC = $AR; 



POSITION OF THE TRUE LINE OF RESISTANCE. 615 

but 

and this is the least distance from the outer edge at which the 
resultant should cut the joint. 

We thus obtain, in terms of the pressure on any joint, and 
of the working-strength of the material, the limits within which 
the line of resistance should pass, in order that the working- 
strength of the stone may not be exceeded. 

§ 262. Position of the True Line of Resistance. — The 
question of the most probable position of the true line of 
resistance involves the discussion of the properties of the 
elastic arch. This discussion will be given later ; but, for the 
present, the statement only of the following proposition, due to 
Dr. Winkler, will be given : — 

" For a?i arch of constant section, that line of resistance is 
approximately the true one which lies nearest to the axis of the 
arch-ring, as determined by the method of least squares." 

From this it will follow : — 

i°. That, if a line of resistance can be drawn in the arch- 
ring, then the true line of resistance will lie in the arch-ring ; 
and 

2°. That, if a line of resistance can be drawn within the 
middle third of the arch-ring, then the true line of resistance 
will lie in the middle third. 

But, before proving this proposition,. the proposition will be 
used, and the method explained, for determining whether a line 
of resistance can be drawn within the arch-ring : for, if it can, 
then the true line of resistance must lie within the arch-ring; 
and if no line of resistance can be drawn within the arch-ring, 
then the true line of resistance cannot pass within the arch- 
ring, and the arch would necessarily be unstable, even if the 
materials were incompressible. 

By following the same method, we could determine whether 



6l6 APPLIED MECHANICS. 

it was possible to draw a line of resistance within the middle 
third of the arch-ring ; and, if this is found to be possible, we 
should know that the true line of resistance will pass within 
the middle third of the arch-ring. 

Hence our most usual criterion of the stability of a stone 
arch is, whether a line of resistance can be passed within the 
middle third of the arch-ring. 

If the condition be used, that the working-strength of the 
stone for compression be not exceeded, then, instead of the 
middle third, we shall have some other limits. 

In what follows, an explanation will be given of Dr. Scheff- 
ler's method (that most commonly employed) of determining 
whether a line of resistance can be drawn within the arch-ring, 
inasmuch as the same method can be employed to determine 
whether such a line can be drawn within the middle third or 
within any other given limits. 

§ 263. Preliminary Proposition referring to Arches Sym- 
metrical in Form and Loading. — An arch and its load being 
given, a line of resistance can always be made to pass through 
any two given points ; hence, if any two points of a line of 
resistance are given, the line is determined. 

Proof. — Let the arch be that shown in Fig. 281 ; and let us 
consider first the special case when the two given points are A, 
the top of the crown-joint, and G A , the foot of the springing- 
joint. In this case, the only quantity to be determined is the 
thrust at A. Let this thrust be denoted by T ; let P be the 
total weight of the half-arch and its load ; let a be the perpen- 
dicular distance of the point c7 4 from a vertical line through the 
centre of gravity of the entire half-arch and its load ; let h be 
the vertical depth of G 4 below A. Then, taking moments about 
£ 4 , we must have 

Th = Pa 

/. T =%; (1) 

h 



DR. SCHEFFLER'S METHOD. 6l? 

and the line of resistance can then be drawn with this thrust, 
as has been done in the figure. Next take the general case, 
when the given points are not in these special positions. Let 
them be any two points, as A 2 and G y 

In this case, the point of application of the thrust at the 
crown is not necessarily A, but may be some other point of the 
crown-joint : hence the quantities to be determined are two ; 
viz., the thrust T at the crown, and the distance x of its point 
of application below A. Let the combined weight of the first 
two voussoirs and their load be P^ and the horizontal distance 
of A 2 from a vertical line through the centre of grav.ity of P 1 be 
a,. 

Let P, be the combined weight of the first three voussoirs 
and their load, and let a 2 be the horizontal distance of G 3 from 
a vertical line through the centre of gravity of P 2 . 

Let the vertical depth of A 2 below A be h 19 and that of G 3 
below A be h 2 . Then, taking moments about A 2 and G 3 respec- 
tively, we shall have 

T{h x — x) == P z a t and T(k 2 — x) = P 2 a 2 , 

two equations to determine the two unknown quantities T and 
x, which can easily be solved in any special case ; and the result- 
ing line of resistance can be drawn, which will pass through the 
two given points. 

§264. Dr. Scheffler's Method of Determining the Possi- 
bility of Passing a Line of Resistance within the Arch- 
Ring. — In using Scheffler's method of determining whether it 
is possible to pass a line of resistance within the arch-ring or 
not, we should proceed as follows ; viz., — 

First pass a line of resistance through 1, the top of the 
crown-joint (Fig. 280), and e, the inside of the springing-joint. 
If this line lies wholly within the arch-ring, it proves that a 
line of resistance can be drawn within the arch-ring. 

If this line of resistance does not pass entirely within the 




6l8 APPLIED MECHANICS. 

arch-ring, proceed as follows-: Suppose the line thus drawn to 
be \abcde, passing without the arch-ring on both sides, as 
shown in the figure. Then from a, the point 
where it is farthest from the extrados of the arch- 
ring, draw a normal to the extrados, and find the 
point where this normal cuts the extrados : in this 
case, a is the point in question. In this way 
determine also the point 7, where the normal 
from d cuts the intrados ; then pass a new line 
of resistance through the points a and 7, deter- 
mining the thrust and its point of application. If this new 
line of resistance lies within the arch-ring, then it is plain that 
it is possible to draw a line of resistance within the arch-ring ; 
if not, it is not at all probable that it is possible to draw such a 
line. 

If the line of resistance drawn through 1 and e goes outside 
the arch-ring only beyond the extrados, as at a, we should draw 
our second line of resistance through a and e ; if, on the other 
hand, it goes outside only below the intrados, as at d> we should 
draw our second line through 1 and 7. 

In the construction, we make use of a slice of the arch in- 
cluded between two vertical planes a unit of distance apart ; and 
we take for our unit of weight the weight of one 'cubic unit of 
the material of the voussoirs, so that the number of units of 
area in any portion of the face of the arch shall represent the 
weight of that portion of the arch. 

We next draw, above the arch, a line. {DD A in Fig. 281), 
straight or curved, such that the area included between any 
portion of it, as DjD 2f the two verticals at the ends of that por- 
tion, and the extrados of the arch-ring, shall represent by its 
area the load upon the portion of the arch immediately below 
it. This line will limit the load itself whenever this is of the 
same material as the voussoirs ; otherwise it will not. We shall 
always call it, however, the extrados of the load. 



DR. SCHEFFLER'S METHOD. 



6. 9 



The mode of procedure will best be made plain by the solu- 
tion of examples ; and two will be taken, in the first of which 
only one trial is necessary to construct a line of resistance that 
shall lie wholly within the arch-ring, and, in the second, two 
trials are necessary. 

Example. — The half-arch under consideration is shown in 
Fig. 281, GG 4 being the intrados, AA 4 the extrados of the arch, 



d 4 d 3 




Fig. 281. 



and DD 4 the extrados of the load. The arcs GG 4 and AA 4 are 
concentric circular arcs. The data are as follows : — 

Span = 2(G 4 0) = 6.00 feet, 

Rise —GO =0.50 foot, 

Thickness of voussoirs = AG — A 4 G 4 =0.75 foot, 

Height of extrados of load above A = AD = 1.60 foot. 



The position of the joints is not assumed to be located. We 
therefore draw through A a horizontal line AB, and divide this 
into lengths nearly equa], unless, as is usual near the springing, 
there is special reason to the contrary. Thus, we make the first 
three lengths each equal to 1 foot, and thus reach a vertical 



620 



APPLIED MECHANICS. 



through G 4 ; and then the last division has a length of 0.24 foot. 
We have thus divided the half-arch and its load into four parts ; 
viz., GDD X H U H.D.D.H,, H 2 D 2 D 3 G 4 , and G 4 D 3 D 4 A 4 , the loads 
on these respective portions being represented by their areas 
respectively. We assume the centre of gravity of each load to 
lie on its middle vertical ; and we then proceed to determine the 
numerical values of the several loads, the distances of their 
centres of gravity from a vertical through the crown, also the 
amount and centre of gravity of the .first and second loads 
together, then of the first, second, and third, etc. 

The work for this purpose is arranged as follows : — 



(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


(8) 


(9) 


(10) 




Length. 


Height. 


Area. 


Lever 
Arm. 


Moment. 


Partial Sums. 


Area. 


Moment. 


Lever 
Arm. 


2 
3 
4 


1. 00 
1. 00 
1. 00 
O.24 


i-57 
1.68 
1.90 

1.72 


I.570 
I.680 
I.900 
O.413 


O.50 

I.50 
2.50 
3.12 


O.785 
2.520 
4750 
I.287 


I 

1 + 2 

1 + 2+3 
1+2+3+4 


I.570 
3-250 
5-I50 
5-563 


O.785 

3-3°5 
8.055 

9-344 


O.500 
I.017 

I-563 
I.680 


- 


3- 2 4 


- 


5-563 


- 


9-344 


- 


- 


- ' 


- 



Column (1) shows the number of the voussoir. 

(2) gives the horizontal lengths of the several trape- 
zoids. 

" (3) gives the middle heights of the trapezoids. 

" (4) gives the areas of the trapezoids, and is obtained by 
multiplying together the numbers in (2) and (3). 

" (5) gives the distances from A to the middle lines of 
the trapezoids. 

" (6) gives the products of (4) and (5), giving the moments 
of the respective loads about an axis through A 
perpendicular to the plane of the paper. 



DR. SCHEFFLER'S METHOD. 621 

Column (7) merely indicates the successive combinations of 
voussoirs. 
" (8) has for its numbers, — 

i°. The area representing the first load. 
2°. The area representing the first two loads. 
3 . The area representing the first three. 
4 . The area representing the first four. 
" (9) has for its numbers, — 

i°. The moment of the first load about A. 
2°. The moment of the first and second loads 

about A. 
3°. The moment of the first, second, and third 

loads about A. 
4 . The moment of the first, second, third, and 
fourth loads about A. 
(10) is obtained by dividing column (9) by column (8) ; 
the quotients being respectively the distance 
from A to the centres of gravity of the first, of 
the first and second, of the first, second, and 
third, and of the first, second, third, and fourth 
loads. 
The calculation thus far is purely mathematical, and merely 
furnishes us with the loads and their points of application ; in 
other words, furnishes us the data with which to begin our 
calculation of the thrust. Before passing to this, it should be 
said, however, that we now assume the joints to be drawn 
through the points A lt A 2} A 3 , and A 4 , and generally normal to 
the extrados of the arch. 

In this proceeding, we, of course, make an error which is 
very small near the crown and increases near the springing of 
the arch ; this error, in the case of voussoir A I A 2 G 1 G 2 , amounts 
to the difference of the two triangles A 2 G 2 H 2 and A.G^. A 
manner of making a correction by moving the joint will be 
explained later ; but now we will complete our example, as the 



622 



APPLIED MECHANICS. 



errors are not serious in this example. We now pass a line of 
resistance through A, the upper point of the crown-joint, and 
C7 4 , the lower point of the springing. For this purpose take 
moments about G 
crown, 



, and we shall have, if T = thrust at the 
i.2 5 r= (5.563) (3 - 1.68), 



since 5.563 is the whole weight, and 3 — 1.68 is its leverage 
about G~ 



Hence 



1.25 r= (5.563X1.32) 

.-. ^=5.87. 



7-343 



Hence we proceed to draw a line of resistance through A, 
assuming, as the horizontal thrust, 5.87. To do this we proceed 
as follows : From a, the point of intersection of AD with the 
vertical through the centre of gravity of the first trapezoid, we 
lay off ab to scale equal to 5.87, and then lay off bC vertically 
to scale equal to 1.57, the first load ; then will Ca be the result- 
ant pressure on joint A X G„ and its point of application will be 
P t which gives us one point in the line of resistance. To 
obtain the point P lt we lay off Aa x = 1.017, the lever arm of 
the first two loads ; then lay off a l b I = 5.87, the thrust ; -then 
lay off b x C x equal to 3.25, the weight of the first two loads. 
Then will C& be the pressure on the second joint ; and the 
point P Jt or its point of application, is at the intersection of 
C x a x with A 2 G 2 . 

Then lay off Aa 2 = 1.563, a 2 b 2 = 5.87, b 2 C 2 = 5.150; and 
P 2 , the next point of the line of resistance, is the intersection 
of C 2 a 2 with A 3 G r Then lay off Aa 3 = 1.680, a 3 b 3 = 5.87, 
b 3 C 3 = 5.563; and C 3 a 3 is the pressure on the springing, and 
this will intersect A 4 G 4 at G 4 unless some mistake has been 
made in the work. Then is APP,P 2 G 4 the line of resistance 
through A and G 4 , and this lies entirely within the arch-ring. 



SCIIEFFLER'S MODE OF CORRECTING THE JOINTS. 623 



Hence we conclude that it is possible to draw a line of resist- 
ance within the arch-ring without having recourse to another 
trial. , 

§ 265. Scheffler's Mode of Correcting the Joints. — The 
following is the approximate construction given 
by Scheffler for correcting the joint : Let DCG 
be the side of the trapezoid, and CH the uncor- 
rected joint. From b, the middle point of GH, 
draw Db ; then draw Gc parallel to bD, and cJi 
parallel to CH. Then will ch be the corrected 
joint. 

Conversely, having given the 

joint CH, to find the side of the trapezoid which 

limits the portion of the load upon it : through 

C draw DG vertical ; join D with b, the middle 

point of GH ; then draw Cg parallel to Db ; 

then, from g, drawing dg vertical, we thus have 

the desired side of the trapezoid. 

§ 266. Another Example. — Another example will now be 

solved, which necessitates two trials, and where some of the 

joints have to be corrected. It is practically one of SchefBer's. 

The dimensions of the arch are as follows : — 





Fig. 283. 



Half-span 3 2 -97 feet- 
Rise 24.74 feet. 

Thickness of ring - 5.15 feet. 

Height of load at crown 8.24 feet. 

Height of load at springing 33-5° feet. 



The arch may be drawn by using, for the intrados, two 
circular arcs. Beginning at the springing, draw a 6o° arc with 
a radius of one-fourth the span ; then, with an arc tangent to 
this arc, continue to the crown, the proper rise having been 
previously laid off. The work for drawing a line of resistance 



624 



APPLIED MECHANICS. 



through the top of the crown-joint and the inside of the 
springing will be given without comment. It is as follows : — 



(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


(8) 


(9) 


(10) 


<L> 3 

§ > 


Length. 


Height. 


Area. 


Lever 
Arm. 


Mo- 
ment. 


Partial Sums. 


Area. 


Moment. 


Lever 
Arm. 


I 


8.2 4 


14.I 


Il6.l8 


4.12 


476 


I 


Il6.l8 


476 


4.12 


2 


8.24 


16.4 


135-14 


12.36 


1675 


1+2 


25I-32 


215I 


8-57 


3 


8.24 


18.3 


15079 


20.6o 


3106 


I + ... + 3 


402.II 


5257 


13.09 


4 


4-^3 


22.6 


93-34 


26.79 


2600 


I+. . . + 4 


495-45 


7857 


I5-87 


5 


4-13 


27.I 


1 1 1.92 


30.92 


3443 


1+...+5 


607.37 


11 305 


18.62 


6 


5-14 


347 


178.36 


35-55 


6351 


i-h ..+6 


785 73 


17656 


22.49 


- 


- 


- 


78573 


- 


17656 


- 


- 


- 


- 



29.89^= (785.73) (32.97 
29.89^= 8234.45 



22.49), 



T= 275.5. 



Hence we construct the line of resistance passing through the 
top of the crown-joint and the inside of the springing, using 
the thrust 275.5. 

The construction is shown in the figure, and is entirely 
similar to that previously used, with the single exception that 
the upper, instead of the lower, half of the rectangle is used, 
in each case, in constructing the parallelogram of forces, to 
determine the pressure on each joint : this is merely a matter 
of convenience. The student will readily identify this first 
line of resistance, and will see that it goes outside the arch- 
ring both above and below, being farthest above the extrados 
at the first joint from the crown, and farthest inside of the 
intrados opposite the first joint from the springing. Hence we 



SCHEFFLER'S MODE OF CORRECTING THE JOINTS. 625 

proceed to pass a new line of resistance through the top of the 
first joint from the crown, and the inside of the first joint from 
the springing. 



/ r-t-p--/^-'" 









i t\ / 


1 1 V / E? 


> + /-- 


/- — -7 


-"--■ 


_-- 


1 


A-fh 
l/\ i 

1 // 




/ /rzt 

\ ^ — 


-£■--"" 




\( 


'i7/ii 








Ijlj* 




I / 


f 





Fig. 284. 

For this purpose we do not need to make out a new table, 
as it is not necessary to insert any new joints. We need only 
two more dimensions, i.e., the vertical depth of each of these 
points below the top of the crown : these depths are respec- 
tively 2.38 and 21.85. 

Hence we proceed as follows : — 
Let T = thrust at the crown, 

x = distance of its point of application below the top of 
the crown-joint. 

i°. Take moments about the upper one of the two points, 
and we have 

7X2.38 - x) = (n6.i8)(8.2 4 - 4.12) = 478.66. 



626 



APPLIED MECHANICS. 



2°. Take moments about the lower one of the two points, 
and we have 

7 T (2i.85 - x) = (607.37) (30.90 - 18.62) = 7458.5°- 
Solving these equations, we obtain 

T = 35 8 -9> x = 1 -°4^- 

Hence we lay off on the crown-joint a distance 1.046 below the 
top of the crown-joint, and through this point draw a horizontal 
line, this line being the line of action of the thrust. Then, 
making the construction for a new line of resistance just as 
before, only using this new point of application of the thrust, 
and using for thrust 358.9, we shall obtain a new line of resist- 
ance, which passes through the desired points ; and, since this 
line lies within the arch-ring, we therefore conclude that it is 
possible to draw a line of resistance within the arch-ring. 

§ 267. Examples. — Four more examples will now be given, 
to be worked out by the student. The dimensions are approx- 
imately those given in some of Schefrler's examples. 

Example I. — Half-span = CD = 65.16 feet, rise = FD = 
13.85 feet, AF = 5.32 feet, AE = 6.40 feet. The arcs CF and 
AG are concentric circular arcs. Given width of first five hori- 
zontal divisions of line AB, counting from A, each 10.66 feet ; 
width of sixth division, 11.84 feet; of seventh, 3.68 feet. De- 
termine the possibility of drawing a line of resistance in the 
arch-ring. 




Fig. 285. 

Example II. — Half-span = 63.98 feet, rise = FD = 31.99 
feet, AF == CG — 5.32 feet, AE =2.13 feet. The intrados and 






EXAMPLES. 



627 



extrados of this arch are seven centred ovals, both drawn from 
the same centres. Beginning at the springing, an arc with a 
radius of 21 feet is drawn, subtending 39 ; the curve is con- 
tinued by a curve subtending 24 , and having a radius of 35-55 
feet. From F an arc subtending io° is drawn from a centre 
on FD produced, and with a radius of 152 feet; the curve is 
completed by an arc connecting the second and last. 




g c 



Fig. 286. 



Given horizontal width of each of first six divisions, counting 
from A, 10.66 feet ; horizontal width of seventh division, 5.32 
feet. Determine the possibility of drawing a line of resistance 



in the arch-ring. 




Fig. 287. 



Example III. — Given span = 74.18 feet; rise = 45.83 



628 



APPLIED MECHANICS. 



feet ; radius of intrados == 82.42 feet ; radius of extrados = 
91.18 feet ; height of load at crown = 8.24 feet ; width of each 
of five divisions nearest crown == 8.24 feet; width of sixth 
stone = 4.13 feet. Determine the possibility of drawing a line 
of resistance within the arch-ring. 

Example IV. — Given span = 37.07 feet ; thickness of 
ring = AB = 3.08 feet ; height of load = BC = 82.42 feet. 
Determine the possibility of drawing a line of resistance within 
the arch-ring. 




Fig. 288. 



§ 268. Criterion of Stability. — It has already been stated, 
that, if a line of resistance can be drawn within the arch-ring, 
then the true line of resistance will lie within the arch-ring. 



UNSYMMETRICAL ARRANGEMENT. 629 

With those who, like Scheffler, consider the material of the 
voussoirs incompressible, the criterion of stability of an arch is, 
that it should be possible to draw aline of resistance within the 
arch-ring. 

On the other hand, Rankine would decide upon the stability 
of an arch by determining whether a line of resistance can be 
drawn within the middle third of the arch-ring. 

Other limits have been adopted instead of the middle third. 
In some cases the only reason for deciding upon what these 
limits should be has been custom or precedent. 

They might also be determined so that there should be no 
danger of exceeding the crushing-strength of the stone. 

It is needless to say that the first method is incorrect ; for 
the material of the voussoirs is never incompressible, and an 
arch where the true line of resistance touches the intrados or 
extrados could not stand, as the stone would be crushed. 

Nevertheless, no example will be solved here, where we de- 
termine the possibility of drawing a line of resistance within 
the middle third, or other limits than the entire arch-ring, as the 
method of procedure is entirely similar to what we have done, 
the computation of the entire table being the same in all cases, 
the only difference occurring in the computation of the thrust 
and its point of application, and the consequent construction of 
the line of resistance. The method to be pursued is, as before, 
by taking moments about the points through which it is desired 
that the line of resistance shall pass. 

§ 269. Unsymmetrical Arrangement. — When the arch is 
unsymmetrical, either in form or loading, the same criterion as 
to being able to pass a line of resistance within the middle third 
or other limits of the arch-ring will serve to determine its sta- 
bility. The method of procedure differs, however, from the fact, 
that whereas we have heretofore found it necessary to study 
only the half-arch and its load; and have had the advantage of 
knowing, from the symmetry of arch and load, that the thrust at 




63O APPLIED MECHANICS. 

the crown is horizontal, we have not that advantage here, and 
hence we must study the entire arch, and we must assume that 
the thrust at the crown may be oblique, and hence have a verti- 
cal as well as a horizontal component. 

In this case it will be necessary to have three instead of two 
points given, in order to determine a line of resistance. 

If we assume (Fig. 289) a vertical joint at the crown, and let 
P = vertical component of the thrust at 
the crown, A = horizontal component of 
the thrust at the crown, x = distance 
of point of application of thrust at the 
crown below upper point of crown-joint, 
we have thus three unknown quantities, 
and we shall therefore, need three equa- 
tions to determine them. 

In this case, therefore, we must have three points of the line 
of resistance given, in order to determine it ; and a reasoning 
similar to that pursued in § 263 would show that a line of re- 
sistance can always be passed through any three given points. 

In performing the work, we should need to make out a table 
for the part of the arch on each side of the crown-joint, show- 
ing the loads, and centres of gravity of the loads, on each vous- 
soir, and on combinations of the first two, first three, etc. ; this 
portion of the work being entirely similar to that done in the 
case of arches of symmetrical form and loading, only that we 
require a separate table for the parts on each side of the crown- 
joint. 

When these two tables have been worked out, we next pro- 
ceed to impose the conditions of equilibrium by taking moments 
about each of the three points given. 

Thus, suppose that (as is usually done first) we pass a line 
of resistance through the top of the crown-joint and the inside 
of each springing-joint, we then* have only two unknown quan- 
tities to determine ; viz., P and Q, inasmuch as x becomes zero. 



UNSYMMETRICAL ARRANGEMENT. 63 I 

Hence we take moments about the inner edge of each of the 
springing-joints. 

In taking moments about the inner edge of the left-hand 
springing- joint, we impose the conditions of equilibrium upon 
the forces acting on that part of the arch that lies to the left of 
the crown-joint. These forces are, (i°) its load and weight, 
which tend to cause right-handed rotation ; (2 ) the horizontal 
component of the thrust exerted by the right-hand portion up07i 
the left-hand portion ; (3 ) the vertical component P of the thrust 
exerted by the right-hand portion upon the left-hand portion. 

It is necessary to adopt some convention, in regard to the 
sign of P, to avoid confusion : and it will be called positive 
when the vertical component of the thrust exerted by the right- 
hand portion on the left-hand portion is upwards ; when the 
reverse is the case, it is negative. 

We next take moments about the inner edge of the right- 
hand springing-joint, and impose the conditions of equilibrium 
upon the forces acting upon the right-hand portion of the arch. 
In doing this, we must observe that we have for these forces, 
(i°) the weight and load which tend to cause left-handed rota- 
tion ; (2 ) the horizontal component Q of the thrust exerted by 
the left-hand portion upon the right-hand portion, — this acts 
towards the right ; (3 ) the vertical component P of the thrust 
exerted by the left-hand portion upon the right-hand portion ; 
and this, when positive, acts downwards. 

Having determined the values of Q and P, we next proceed 
to draw the line of resistance; and this is done in a similar way 
to that pursued with symmetrical arches, only that the thrust, 
i.e., the resultant of P and Q, is now oblique, and that it acts in 
opposite directions on the two sides of the crown-joint. 

• Having drawn this line of resistance, if we find that it passes 
outside of the arch-ring, we draw normals through the points 
where it is farthest from the arch-ring, and thus obtain three 
points through which to draw a line of resistance : then, taking 



632 



APPLIED MECHANICS. 



moments about each of these three points, we determine, from 
the three resulting equations, values of Q, P, and x, and pro- 
ceed to draw our new line of resistance ; and, if this does not 
pass entirely within the arch-ring, it is not at all probable that 
a line of resistance can be drawn within the arch-ring. All the 
above will be made clearer by the following example : — 

Example. — Given an unsymmetrical circular arch, shown 
in the figure, the intrados and extrados being concentric circles, 
AB = 4', AC = i'.85, MH = i'.5, AH = o'.5, AK == o'.8, to 




Fig. 290, 



determine the possibility of drawing a line of resistance in the 
arch-ring. The tables following show the mode of dividing up 
the load, and getting the centres of gravity, also the mode of 
arranging the work for this purpose. 



UNSYMMETRICAL ARRANGEMENT. 



633 



LEFT-HAND PORTION. 



. 
1- .b 

y 

E 3 
= 
2> 


Width. 


Height. 


Area. 


Lever 
Arm. 


Moment. 


Partial 
Sums. 


Area. 


Moment. 


Lever 
Arm. 


I 
2 

3 
4 
5 


I. OO 
I. OO 
1.00 

1.00 

o-33 


I.32 

I.48 
I.84 
2.42 
2.63 


I.32 
I.48 

I.84 
2.42 
O.87 


O.50 
I.50 

2.50 

3-50 
4.17 


O.660 
2.220 
4.600 
8.470 
3.628 


I 

I + 2 

1 + 2 + 3 
i + ...+ 4 
1 +...+ 5 


I.32 

2.80 
4.64 
7.06 

7-93 


O.660 
2.88o 

7.480 

1 5-950 
I9-578 


O.50 
I.03 
I.6l 
2.26 

2.47 


- 


- 


- 


7-93 


- 


I9-578 


- 


- 


- 


- 



RIGHT-HAND PORTION. 



. 

fl 

- 


Width. 


Height. 


Area. 


Lever 
Arm. 


Moment. 


Partial 
Sums. 


Area. 


Moment. 


Lever 

Arm. 


I 

2 


1. 00 

1. 00 


I.32 

I.48 


I.32 

I.48 


O.50 
I.50 


O.660 
2.220 


I 

I + 2 


I.32 

2.80 


O.660 
2.880 


O.50 
I.03 


- 


- 


2.8o 


- 


2.880 


- 


- 


- 


- 



Now take moments about the left-hand springing inner 
edge, and we have 

2.02Q + $P= 7-93(4 - 2.47) = 12.1239. 

Then take moments about the right-hand springing inner edge, 
and we have 

°-79<2 + i.86/ > = 2.80(1.86 — 1.33) = 2.3240. 



Solving these two equations gives us 

<2 = 4-602, 
P = 0.707. 



634 APPLIED MECHANICS. 



If R represent the resultant of P and Q, we have 



R = \/P 2 + <2 2 = 4.66 ; 

hence we proceed, as follows, to pass a line of resistance 
through the top of the crown-joint and the inner edge of 
each springing: — 

Through the top of the crown draw a horizontal line BAC. 
Lay off Aa = 4.602 and ab = 0.707, and draw y2£/ then 
Ab = 4.66 represents, in direction and magnitude, the thrust 
at the crown. Using this thrust in the same way as we did the 
horizontal thrust in the case of symmetrical arches, we obtain 
the line of resistance EdAF, which is farthest outside of the 
arch at d ; hence, drawing a normal to the arch from d, we ob- 
tain c, the upper edgQ of the first joint from the crown. Hence 
we proceed to pass a new line of resistance through E, c y and F. 

To do this we must assume Q, P, and x all unknown. 

i°. Take moments about P, and we have 

(2.02 — x)Q + 4P = 12.1239. 

2°. Take moments about F, and we have 
(0.79 — x)Q — 1.86P = 2.3240. 

3 . Take, moments about c, and we have 

(0.82 - x)Q -f P= (1.32) (0.50) = 0.66. 

Solving these three equations, we obtain 

Q= 4-9 2 > 
P = 0.64, 
x = 0.078. 
Hence 



R = \JP 2 + Q 2 = 4-96. 

Hence, if we lay off a distance 0.078 below A, we shall have 
the point on the crown-joint at which the thrust is applied; 



GENERAL REMARKS. 635 

and making the same kind of construction as we just made, 
only using this point instead of A, and these new values of Q 
and P, we construct the second line of resistance. The con- 
struction is omitted in order not to confuse the figure ; but the 
line of resistance is drawn, and the student can easily make 
the construction for himself. It will be seen, that, in this 
case, this new line of resistance lies entirely within the arch- 
ring. 

§ 270. General Remarks. — Whenever there are also hor- 
izontal external forces acting upon the arch, these should be 
taken into account in imposing the conditions of equilibrium. 

It will be noticed, that, in the preceding discussion, it has 
always been assumed that the load upon any one voussoir is the 
weight of the material directly over that voussoir. This is the 
assumption usually made in computing bridge arches : and it 
may be nearly true when the height of the load above the crown 
is not great ; but even then it is not strictly true, and when 
this depth becomes great, as would be the case with an arch 
which supports the wall of a building, it is far from true, as the 
distribution of the load actually coming upon different parts of 
the arch must vary with, and depend upon, the bonding of the 
masonry, and also upon the co-efficient of friction of the mate- 
rial. Thus, in the case of an arch supporting a part of the wall 
of a building, it is probable that the only part of the load that 
comes upon the arch is a small triangular-shaped piece directly 
over the arch, and that above this the material of the wall is 
supported independently of the arch. This will be plain when 
we consider, that, were such an arch removed, the wall would 
remain standing, only a few of the bricks near the arch falling 
down ; and though the number of bricks that would fall would 
be greater while the mortar is green, still even then only a few 
would drop out. 

In regard to these matters, we need experiments ; but thus 
far we have none that are reliable. 



636 APPLIED MECHANICS. 

Then, again, we have arches supporting a mass of sand or 
gravel ; and then the mutual friction of the particles on each 
other comes into play, and it is not true in this case that the 
load on any voussoir is the weight of the material directly 
above that voussoir. In some cases this has been accounted 
as a mass of water pressing normally upon the arch, but we 
cannot assert that such a course is correct. 

On the other hand, there are cases where we know that an 
arch is subjected to horizontal as well as to vertical forces, and 
sometimes we cannot tell how great these horizontal forces are. 
Thus, the forms of sewers are an arch for the top and an 
inverted arch for the bottom ; but in this case the sides of the 
ditch in which the sewer is laid when building it, are capable of 
furnishing whatever horizontal thrust is needed to force the 
line of resistance into the arch-ring, provided that a horizontal 
thrust is what is needed to force it in. Hence it is, that, were 
the attempt made to pass a line of resistance within the arch- 
ring of almost any successful sewer, accounting the load as the 
weight of the earth above it, the line would almost invariably 
go outside ; but the earth on the sides is capable of furnishing 
the necessary horizontal thrust to force it inside, unless a care- 
less workman has omitted to ram it tight, or unless some other 
cause has loosened it on the sides of the sewer. 

If we know, in any case, the actual law of the distribution 
of the load, we can determine the proper form for the arch by 
the methods of the first part of this chapter, as was done in 
the case of the parabola and of the catenary. Scheffler's 
method is, however, the one almost always used for determin- 
ing the stability of any stone arch against overturning around 
the joints. 

Should there ever arise a case where there was danger that 
the resultant pressure on any joint made an angle with the 
joint greater than the angle of friction, this could be remedied 
by merely changing the inclination of the joint. 




GENERAL THEORY OF THE ELASTIC ARCH 637 



§ 271. General Theory of the Elastic Arch. — In the case 
of the iron arch, the loads upon the arch are all definitely 
known ; and it is necessary to ascertain with certainty the stress 
in all parts of the structure, and to so proportion the different 
members as to bear with safety their respective stresses. 

The general discussion of the method used in calculating 
such arches will now be given ; the method used being practi- 
cally that followed by Dr. Jacob J. Weyrauch, and explained 
more at length in his "Theorie der Elastigen Bogentrager." 

This discussion is also necessary in order to prove the 
proposition already enunciated in § 262 ; viz., that "for an arch 
of constant cross-section, that line of resistance is approximately 
the true.one which lies nearest to the axis of the arch-ring as 
determined by the method of least squares." 

In this discussion the following definitions are adopted : — 

i°. The axis of the arch is a plane curved line passing 
through the centres of gravity of all its normal sections. 

2°. The plane of this axis is called the plane of the arch. 

3°. The axial layer of the arch is a cylindrical surface per- 
pendicular to the plane of the arch, and containing its axis. 

4 . A section normal to the axis is called a cross-section. 

5°. The length of the axis between two sections is called 
the length of arch between the sections. 

The loads may be single isolated loads, or they may be 
distributed loads. 

We shall, in this discussion, assume in the plane of the arch 
a pair of rectangular axes, OX and O Y, positive to the right and 
upwards respectively. 

We will, then, assuming any point on the axis of the arch 
before the loads are applied, call x y y, the co-ordinates of that 
point, s the length of axis from some arbitrary fixed point, <£ the 
angle made by the tangent line at that point with OX, r the 
radius of curvature of the axis at that point, x -f- dx, y + dy, 
s -\- ds, and <f> -j- dcj>, the corresponding quantities for a point 



633 



APPLIED MECHANICS. 




very near the first before the load is applied ; also we will denote 

by 77 the perpendicular distance of 
any fibre from the axial layer, by S^ 
the length of arc measured to that 
point where this fibre cuts the cross- 
section through (x, y), and S v -f- 
dS v the length of arc measured on 

this fibre to the next cross-section, 

FlG 2 9 J - so that ds will be the distance apart 

of the cross-sections measured on the axis, and dS^ on the other 

fibre. All this is done before the 

load is applied, and is shown in 

Fig. 291 ; while the changes brought 

about by the application of the 

loads are denoted by A's, and 

shown in Fig. 292. Thus, x, y, s, 

and <f> become respectively x -j- 

Ax, y + Ay, s + As, and <f> -f- A<£. 
Now, the course we are to fol- 



Y 


1 
1 
1 
1 | 

Adx\ ^? | , 


*Ay> 




y\ 

z\ ! 1 ! 


X 



Fig. 292. 



low in the discussion is, 

a cross-section dividing the arch 

into two parts, and to impose the conditions of equilibrium 

between the external forces acting on the part to one side of 

the section, and the forces exerted by the other part upon this 

part at the section. These latter forces may be reduced to the 

three following : — 

i°. A normal thrust T x uniformly distributed over the sec- 
tion, the resultant acting at the centre of gravity of the section. 

2°. A shearing-force S x at the section. 

3 . A bending-couple at the section ; this comprising a 
stress varying uniformly from the axial layer, and amounting 
to a statical couple, tension below, and compressions above, the 
axial layer. 

Moreover, (1) and (3) combined amount to a uniformly vary- 



GENERAL THEORY OF THE ELASTIC ARCH. 639 

ing stress, the magnitude of whose resultant is T X1 its point of 
application not being at the centre of gravity of the section ; 
this sort of composition having been already exhibited in the 
case of the short strut (§ 207). 

Now, let r be the radius of curvature of the axial layer at 
the section ; and we have, from Fig. 291, by similar sectors, 

dS, = dS ■+ r}(-tt<(>) = dS — rjd<l>. (1) 

But 

r(-<ty) = ds .-. * - -I 

ds r 



ds v 



=4 + ;M^> (2) 



Now, if the loads are applied, and the changes take place 
that are indicated in Fig. 292, we shall have, by suitable sub- 
stitutions in (1), 

d(S v + AS,) = d(S + As) - v d(4> + A</>); (3) 

and, combining this with (1) and (2), we obtain 

dAS n fdAS dAc{>\ r 

~dS~, = \ds ~ V ~ds~) F~Vrj ^ 

Now, the change of length of fibre from dS, to d(S v + AS,) is 
due to two causes : (1) the change of temperature, (2) the stress 
acting on the fibre normal to the section. 

Let e = co-efficient of expansion per degree temperature, 
r = difference of temperature, in degrees. 
p v = intensity of stress along the fibre at section. 
E = modulus of elasticity of the material. 
Then 

€r E~~ dSr, ~~ \ds V ds )r + rj ' 5) 

Hence, solving for p vy we have 

/ dAd> dAs\ r 

A-^-.-JTrT,^ (6) 



64O APPLIED MECHANICS. 

this being the expression for the stress per square inch on the 

fibre whose distance is 77 from the axial layer. 

Hence we shall have, by summation, if elementary area = 

dA, 

fd^.6 rr\dA d&s rdA ~] 

T x = 3*fA = ^t^- - -^2— + Add], ( 7 ) 

and for the moment M x we have, by taking moments about the 
neutral axis of the section (i.e., horizontal line through its cen- 
tre of gravity), 

Vd\&^rrfdA d\s^rr,dA ~] , , 

Let tdA — A, 2 = fi, and observe that ^dA = o, 

r + n 
since the axis passes through the centre of gravity of the sec- 
tion, and we have 

rrjdA rfdA 

r -f 7] r + 77 

rdA 1 1 ^, rfdA O 



r + r) r r r + rj r 

Making these substitutions, we have 



E 



\ ds r ds J r \ds J ' 

/ d£<f> dAs\Q 

\ r ~dT + ~diJ7 



M x I d\<b d±s\Q 



Hence, solving for and — -?, we have 

ds ds 



d\s 
ds 



I M x 1 \ 



(9) 



dAcf> / M x \ 1 M x er v . . 

W = \ T * + -7)EAr + m--r= X - < xo > 



GENERAL THEORY OF THE ELASTIC ARCH. 64 1 

Now, from Fig. 292, we have 

d(x -f Ax) = d(s -f- As) cos (</> -f A<f>), 
d(y + Ay) = d(s + As) sin (</> + Ac/)); 

but, if we write cos A</> = 1, and sin A<f> =■ Ac/>, 

v . dx . dy 

cos(c/> ■+- Ac/>) = cosc/> — A<£smc/> = — — A</>— » 



Hence 



.... dy dx 

sm(c/> + A</>) = sin c/> + A</> cos c/> = -=■ + A<f>-r* 

dAx = —Acjidy -\ */x — f ^Ad> ), 

*& \ ds " J 

dAy = +A$dx + — /#> + f—dxAcf>\. 
ds \ ds J 

or, omitting the last terms, and integrating, 

Ax = -fA$dy + /Kfc, (n) 

Ay = fA^dx+fYdy; (12) 

and, integrating (9) and (10), 

A* = /Kfr, (13) 

Ac/) = /^ifc. (14) 

In these four equations we have 

Ax = horizontal deflection due to the loads, 
Ay = vertical deflection due to the loads, 
As = change of length of arc due to the loads, 
Ac/> = change of slope due to the loads. 

If, now, we write 



M x 


= 


EQ, 


+ 


EAr* 


+ 


EAr 


/>! 


= 


M x 
EAr 


+ 


T x 
EA' 







642 APPLIED MECHANICS. 



we shall have 



r ' 



and hence (n) to (14) become 

Ax = -f/M.dsdy - fP L dx 4- / / -dsdy - ferdx, (15) 

Ay = f/M.dsdx - fP.dy - j j ~dsdy - ferdy, (16) 
Ai" .== - fPJs - ferds, (17) 

A<f> = /j/;^ - / -<&. (18) 

If we neglect the effect of temperature, they become 

Ax = -ffM x dsdy - fPJx, (19) 

Ay = f/MJsdx - fPJy, (20) 

aj = - yvvft, (21) 

Ac£ = fM x ds. (22) 

If, on the other hand, we do not neglect the effect of tempera- 
ture, but omit all terms containing - in the values of X and 

Y, which would be more nearly correct the larger the value of 
r, i.e., the flatter :tbe arch, we should obtain 

Ax = ~ J imi dsdy ~ §7TA dx - ~ f " dx ' (23) 

^ = JSm dsdx ~ H dy ~ s " dy ' (24) 

te = - J -£?-/* - f"ds, (25) 

A * = jf is*- < 26 > 



DETERMINATION OF STRESSES IN AN IRON ARCH. 643 



Moreover, if we put the moment of inertia / for ft, which will 
cause but little error, we derive 

Ax = ~ S'fm** ~ Im dx ~ JiTdx > < 2 ?) 
*- //£**. -/S*. ~^"* (28) 

As = - J £2 ds ~ S" ds > ( 2 9> 

*+= J ju*- (30) 

§ 272. Manner of using the Fundamental Equations to 
Determine the Stresses in an Iron Arch. — In order to be 
able to determine the stresses in all the members of an iron 
arch with any given loading, we need to determine the three 
quantities T m S x , and M x for each section. 

Now, if we let R x represent the thrust at the section, we 
shall have 

R x = ^Tj~TsT\ (i) 

and, if we let H x and V x represent the horizontal and vertical 
components of R x respectively, we have that we need to deter- 
mine the three quantities H x , V xi and M x for each section. 

If' we suppose the arch to be subjected to vertical loads 
only, we shall have, if we let 

H = horizontal component of thrust at all points, 

V — vertical component of left-hand support, 

V I = vertical component of right-hand support, 
M = bending-moment at left-hand support, 

M' =z bending-moment at right-hand support. 

Assume origin of co-ordinates at left-hand support, and 
x + to the right, and y -f- upwards, and impose the conditions 
of equilibrium upon the forces acting on the part of the arch 



644 APPLIED MECHANICS. 

between the section and the left-hand support ; then we have, 
if J^is any one load, and a the x of its point of application, 

H x = H, (2) 

V x = V~^W, (3) 

M x = M + Vx -By - X x W(x - a). ( 4 ) 

Hence it is plain that the three quantities which we need 
to determine are H y V, and M. 

Now these are also the three unknown quantities which will, 
by suitable reductions, become the three unknown constant 
quantities in equations (27) to (30). The determination of these 
three quantities requires three conditions ; what these condi- 
tions are depends upon the manner of building the arch, as will 
be seen from the following three special cases : — 

Case I. — Let the arch be jointed at three points, viz., the 
two supports, and one other point whose co-ordinates are x ss 
x x and y = y x . Then we know, that, for all points where 
there is a hinge, there can be no bending-moment. Hence 

M — o, M r = o, and M Xi = o, 

which are the three required conditions ; and, if these be im- 
posed, it is easy to obtain H xt V x> and M x for every section. 

Case II. — Let the arch be jointed only at the ends. Then 
M = M' = o gives us two conditions : and for the third we 
have A/ = o; i.e., if we put / for x in equation (15) or (27), 
§ 271, after having made the integrations, we have the third 
equation, as this expresses simply the condition that the sup- 
ports remain at the same horizontal distance apart after the 
load is put on as before. With these three conditions we can 
determine H x , V x , and M x for all sections. 

Case III. — Let the arch be fixed in direction at the ends. 
We must now have three conditions. These will be as follows : — 

i°. A/ = o; i.e., the supports remain at the same horizontal 
distance apart after the load is applied as before. 



DETERMINATION OF STRESSES IN AN IRON ARCH. 645 

2°. Ah = o ; {h being the difference of level of the sup- 
ports) ; i.e., the supports remain at the same vertical distance 
apart after as before the load is applied. 

3°. Ac£, = o ; i.e., the tangents at the ends make the same 
angle with each other after as before the load is applied. 

The value of A/ is obtained by integrating (15) or (27), 
§ 271, and then substituting / for x. 

The value of Ah is obtained by integrating (16) or (2%), 
§ 271, and then substituting / for x t or h for y. 

The value of A<£ r is obtained by integrating (18) or (30), 
§ 271, and then substituting / for x, or h for y. 

In this case we often find it convenient to use these equa- 
tions in a simpler form. Thus, suppose we adopt the case 
where we neglect the effect of temperature (i.e., equations (19), 
(20), and (22) of § 271), by making one integration they become 

A* = -JM x yds - fPJx, (5) 

Ay = +/M l x<?s - fPJy, (6) 

A$=fM y ds; (7) 

these being more convenient to use. 

For very flat arches, P x becomes very small : and then the 
equations become 

Ax = -fM.yds, (8) 

Ay = +/M s xds, (9) 

Acf> = fMJs; (10) 

and these, if all terms containing - be omitted, become 






646 APPLIED MECHANICS. 



EXAMPLES. 

i. Given a semicircular arch jointed at each springing-joint and at 
the crown, radius r* Trace out the effect of a single load W acting 
upon it at the extremity of a radius making 45 ° with the horizontal. 

Solution. 

The presence of three joints gives us the bending-moments at each 
of these joints equal to zero, the co-ordinates of these joints being 
respectively (o, o), (r, r), and (2;-, o). 

Hence, using equation (4), we obtain 

i°. M = o, 

2 . Vr — Hr — Wio.^o'jwr) — o, 

3 . V{2r) — ^(1.7071 ir) = o. 

Solving, we have, therefore, 

V — 0.85355 W = left-hand supporting-force, 
and 

H — 0.14645^ = horizontal component of thrust. 

Hence V l = 0.14645 W 7 = right-hand supporting-force. 
Hence, for a section whose co-ordinates are {x x y), 

x < 0.29289;% V x — 0.85355 W ; 

x > 0.29289?-, V x = — 0.14645 W. 

Hence equation (1) gives, for 



x < 0.292897-, R x = ^V(o.85355) 2 -f- (0.14645 ) 2 

= 0.86602 W, 



x > 0.29289/-, R x = W\ (0.14645 )~ -f- (0.14645) 2 

= 0.2071 1 £F. 

Now, the angle made by R x with the horizontal is, for 

/o. 85355V 
x < 0.29289/-, a, = tan- 1 ^ J = 8o° 15' 51", 

x> 0.2 9 2S 9 r, a, = ta-^-_^ = 45 . 



TRUE LINE OF RESISTANCE IN A STONE ARCH. 647 

Knowing, now, the angle made by R x with the horizontal, we can 
find, for the point (x, y), the angle made by a tangent to the circle with 

/r — x\ 
the horizon, or a 2 = tan- z ( ). Then resolve R x into two com- 
ponents, respectively tangent to the arch and normal to it at the point 
x, y, and the tangential component is the direct thrust T X) while the 
normal is the shearing-force S x . 

Then, for the bending moment, we have, from (4), 

_r< 0.29289?-, M x = 0.85355 Wx — 0.14645 Wy ; 

x > 0.29289?-, M x — 0.85355 Wx — 0.14645 Wy — W(x — 0.29289?-). 

Hence we determine the direct thrust, the shearing-force, and the 
bending-moment at any section, and can hence obtain the stresses at all 
points. 

2. Given the same arch with a load W distributed uniformly over 
the circular arc, find stresses at all points. 

3. Given the same arch jointed only at the two springing-points, 
find stresses at all points. 

§ 273. Position of True Line of Resistance in a Stone 
Arch. — The proof will now be given of the proposition already 
referred to in regard to the position of the true line of resist- 
ance ; viz., — 

" For an arch of constant section, that line of resistance is 
approximately the true one which lies nearest to the axis of the 
arch-ring, as determined by the method of least squares." 

Proof. — If we denote by y the ordinate of the axis of the 
arch for an abscissa x, and by /* that of the line of resistance 
for the same abscissa, then /*, — y is the vertical distance be- 
tween the two curves for abscissa x. Now, the condition that 
the line of resistance should be as near the arch-ring as possi- 
ble, is, that the sum of the (//. — y) z shall be a minimum, or 

Sip ~ y)*ds = minimum. (1) 

But (x, fx) are the co-ordinates of the point of application of the 



648 APPLIED MECHANICS. 

actual thrust, and hence (/x — y) is the distance of the point at 
which the resultant thrust acts from the centre of gravity of 
the section. Hence we have 

M x 

Hence (1) becomes 



C(M X \* 

I ( -— ) as = minimum. (2) 



But H is constant for the same line of resistance, though it 
varies for different lines : hence we can place H outside of the 
integral sign. Hence we may write 

u = — - J M x 2 ds == minimum. (3) 

Now, from (4), § 272, we have 



M x = M + Vx - By — *l*W(x - a) = <f>(M, V, &); 

M, V, and H being constants for the same line of resistance, 
but varying for different lines. Hence, by differentiating (3), 



we have 
du du dM v 



dM dM x dM 

du _ du dM x 
~dV~ aW x ~dV 

du du dM, 



dH dM x dH 



= W 2 J Mxds = ° V yMJs^o, (4) 

= ~ \MjcXds = o .*. XM^ds — o (5) 

" == - 2^7-3 Cm x *Os + 2H~ 2 CM x yds 

= -B-^ 2 JMjds +2JjJM x yds}^ = o. 

But the first term must be very small : hence we may write ap- 
proximately, 

fM x yds = o. (6) 

Now, the three expressions (4), (5), and (6) are identical with 
(11), (12), and (13) of § 272 ; and the conditions that these shall 
be zero are, as will be seen by referring to § 272, Case III., the 



DOMES. 



649 



conditions that hold in the case of an arch fixed in direction at 
the ends. Hence it follows that the condition that the line of 
resistance shall fall as near the centre of the arch as possible is 
the condition which, in an elastic arch fixed in direction at the 
ends, gives us its true position. Hence it would seem that 
the most probable position for the true line of resistance is the 
nearest possible to the axis of the arch. 

This is the conclusion reached by Winkler ; and a more 
detailed discussion of the matter is to be found in an article 
by Professor Swain in "Van Nostrand's " for October, 1880. 

§ 274. Domes. — The method to be used for determining 
the stability of a dome differs essentially from that used in the 
case of an arch, for there is no thrust at the crown in a dome. 
Indeed, the most general case is that of the dome open at the 
top : we will, therefore, consider this case first in studying the 
action of the forces required to preserve equilibrium. 

Fig. 243 shows a meridional section of an open dome, 
pose that this dome had been 



Sup- 




entirely built, except the up- 
per ring-course of stones, rep- 
resented by LKGH. Then, 
suppose that one of the stones 
only of this course were placed 
in position without any auxil- 
iary support, its own weight 
would evidently overturn it, 
since the line ad, along which 
the weight acts, does not cut 
the joint ; but, if the whole 
ring-course is put in place, the 
stones keep each other in po- 
sition. The way in which this 
is accomplished is as follows : 
they press laterally against eacli other; and the resultant of the 



\ 



Fig, 293. 



65O APPLIED MECHANICS. 

pressures exerted upon the two lateral faces of any one stone 
by the other stones of the course is a horizontal radial force, 
which, combined with the weight of the stone, gives, as the 
resultant of the two, a force which cuts the joint between G 
and H Moreover, sufficient pressure will be developed to 
accomplish this result, as a failure to reach the result will only 
increase the pressure upon the lateral faces. 

Moreover, if, when sufficient pressure has been developed 
to bring the resultant of the weight of the stone and the above- 
described horizontal radial force within the joint, it should 
make an angle with the normal to the joint greater than the 
angle of friction, the tendency of the stone to slide will increase 
the lateral pressure, and this in turn will increase the outward 
horizontal force till the angle made by the resultant with the 
normal to the joint is no greater than the angle of friction of 
the material of the voussoirs. 

This will be made plain by reference to the figure (Fig. 
243), where ab represents the weight of the stone HLKG, and 
where 0$ is perpendicular to HG and Oy is drawn so that yOS 
= (f>, the angle of friction. Now, since ab produced passes out- 
side of HG, horizontal thrust must be developed. And, more- 
over, were only sufficient horizontal thrust furnished to make 
the resultant cut HG at G, the angle between this resultant 
and the normal to the joint would be greater than<£; there- 
fore we proceed as follows : assuming the horizontal thrust to 
act through Z, the upper edge of the stone, we lay off from b, 
the intersection of the horizontal through Z with a vertical line 
drawn through the centre of gravity of the stone, the weight ab 
to scale, then from b draw be parallel to Oy, and draw through 
a a horizontal line to meet be. Then will ac be the horizontal 
force that will be furnished by the other stones of the course 
to keep this stone in place ; and the pressure upon joint HG 
is be, and acts at the intersection of be and HG. 

Now prolong be to meet the vertical drawn through the 



DOMES. 65 



centre of gravity of the next stone, HGFE, at d. Combine it 
with the weight of this stone ; this is done by laying off de — be, 
and from e drawing ef vertical, and equal to the weight of 
FHGE. The resultant fd makes an angle with the normal to 
FE greater than <f> : hence draw 08 perpendicular to FE and 
Oe, so that c(9S = <£ ; then from g, the intersection of 4^ with a 
horizontal line through H, the top of FHGE, lay off gs = df> 
through g draw gh parallel to <9e, and through s draw sh hori- 
zontal. Then is sh the horizontal thrust that will be furnished 
at H to keep the stone HGEF in place ; and this is the pres- 
sure upon joint FE, and acts at the intersection of FE with kg. 

Next, prolong kg to meet the vertical through- the centre of 
gravity of stone FEDC at k; lay off kl.= gh, and from / lay 
off Im '■==■ weight of stone FEDC; draw km, which cuts the 
joint within the joint itself, and needs no horizontal thrust to 
bring it inside ; hence ink is the pressure on joint DC. 

Then draw mk to meet the vertical through the centre of 
gravity of ABCD at n, and lay off no — km; draw op == weight 
of ABCD, and draw pn, which will be the pressure on the joint 
BA. 

It is necessary, for stability, that all these forces should cut 
the joint inside of the joint if the stones are reckoned incom- 
pressible ; or we may adopt the middle third, or other limits, as 
our criterion of stability. 

As long as it is outward thrust that is required to produce 
stability, it is possible to furnish it ; but, if we should reach a 
joint where inward thrust would be required, this could not be 
furnished, and the dome would be unstable. Moreover, the 
resultant pressure on the springing gives us the pressure ex- 
erted upon the support of the dome ; and it must not cut any 
joint of the support outside of that joint, as otherwise the sup- 
port would not stand. 

In determining the numerical value and direction of this 
pressure on the support, we may either construct it graphically, 



652 APPLIED MECHANICS. 

or we may compute it as follows : (i°) Compound all the ver- 
tical forces, i.e., the weights, and find the magnitude and line 
of action of the resultant of these. (2 ) Compound all the 
horizontal forces, and find the magnitude and line of action of 
their resultant (in this case the horizontal forces are two ; viz., 
ac applied at L, and sh applied at H) ; then compound these two 
resultants. The graphical and analytical method should check 
if no mistake has been made in the work. 

In the above calculation, it has been assumed that the figure 
represents the portion of a dome included between two merid- 
ional planes. 

If we desire to ascertain the pressure exerted upon the 
lateral face of the stone by its neighbors in the same ring- 
course, we only need to know the angle made by the two 
meridional planes containing the lateral faces of the stone in 
question, then resolve the horizontal thrust upon that stone 
into two equal components, which make with each other an 
angle equal to the supplement of the angle of the planes ; i.e., 
resolve the outward horizontal thrust into two components 
normal to the lateral faces. 

In regard to the assumption that the outward thrust acts at 
the top of the stone, it should be said that this is Scheffler's 
custom, his reason being that less thrust will be required if he 
assumes it at the top than if he assumes it nearer the middle. 
The true position of this thrust is probably much nearer the 
middle of the stone. 

An example will next be solved, giving SchefHer's method 
of working. 

Example. — Given the dome shown in the figure, sur- 
mounted by a lantern at the top ; determine whether it is 
stable, and what should be the thickness of the support in order 
that the resultant pressure may not pass outside any joint of 
the pier. 

The dimensions are as follows : — 



DOMES. 



653 



— -JB 



Diameter of outer vertical circle = 20 feet. 

Diameter of inner vertical circle = 18 feet. 

Angle made by springing-radius with vertical = 75 
angle A OB. 

The inner edge of the upper 
voussior subtends 18 on the 
lower circle ; the width of the 
load of the lantern is 0.6 ; 
the voussoirs below that, each 
subtend 18 . 

Assume 36 stones in a hori- 
zontal course. The width of the 
lowest will, then, be 1.51 ; the 
width of the others are deter- 
mined from their lever arms. 

Given height of pier = 8 
feet. 

Height of the centre of the 
sphere above base of pier = 8' 
— iosin 15° = 5.41'. 

The figure may be taken to 
represent the portion of the 
dome included between two ver- 
tical planes passing through the 
axis of the dome : hence it 
shows one vertical series of 
stones. 

We first construct a table 
giving the weights of the differ- 
ent voussoirs with any superin- 
cumbent load, their centres of gravity, and the moments of 
their weights about an axis passing through O, and perpen- 
dicular to the central plane of the portion shown ; and we 
so choose our unit of weight that the volumes of the 




654 



APPLIED MECHANICS. 



voussoirs shall represent their weights, 
as follows : — 



The work is arranged 



Elementary Forces. 


Horizontal Forces. 


(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


(8) 


(9) 


'8 

*s 1 


Area of 
Lateral Face. 


Thick- 

. ness. 


Product. 


Lever 

Arms. 


Moment. 


Hori- 
zontal 
Forces. 


Lever 
Arms. 


Moment. 


i 

2 

3 
4 


0.6X6.680 
2.985 
2.985 
2.985 


o,53 
0.74 
1.23 
i-5i 


2.124 
2.209 
3.672 

4-5°7 


3-07 
4-75 
7-05 
8.68 


6.52I 
IO.493 
25.888 
39.121 


I.74 
I.26 
I.32 


9.60 

9-33 

7.78 


16.IO4 
II.756 
IO.273 


- 


- 


- 


12.512 


- 


82.023 


4-32 


- 


3S730 



Column (1) contains the numbers of the voussoirs, counting 
from the top. 

Column (2) contains the areas of the lateral faces of the 
stones shown in the figure. For the three lower stones, the 
area of a ring subtending 18 at the centre, and of the dimen- 
sions given, is calculated. For the first, the height is 6.68 and 
the width 0.6. 

Column (3) contains the thicknesses of the voussoirs ; i.e., 
the length of arc between their two lateral faces measured on a 
horizontal circle through the centre of gravity of the voussoir, 
which is here taken at the middle point of the arc subtended 
by this voussoir on its middle vertical circle, i.e., one which 
has a radius 9.5 feet. 

Hence, the thickness of the lower stone being 1.51 feet, that 
of the others will be 



(i - 5i) l3 = °- 53 ' 



(1.50 



7-05 

8.68 



/ \4-25 



1.23. 



DOMES. 655 



Column (4) gives the weights of the voussoirs and their 
loads : it is obtained by multiplying together the numbers in 
columns (2) and (3). 

Column (5} gives the distances of the centres of gravity of 
the different voussoirs from the axis of the dome : it may be 
determined graphically or by calculation. 

Column (6) gives the moments of the weights about a hori- 
zontal axis through O perpendicular to the central plane of this 
series of voussoirs. The graphical construction for determin- 
ing the horizontal thrusts required is next made, and the results 
are recorded in column (7). It will be seen that no thrust is 
required on voussoir No. 4. 

Column (8) contains the lever arms of these forces about 
the same axis. 

Column (9) contains their moments about the same axis. 

The construction thus far has shown no case where horizon- 
tal tension instead of horizontal thrust is required to cause the 
thrust on any joint to pass within the joint : hence thus far the 
dome is stable ; and the question comes next as to what should 
be the width of the pier in order that the line of resistance, if 
continued down, may remain within it. 

For this purpose we proceed as follows : — 

Let t = thickness required. 

Let breadth be equal to that of the lowest voussoir. 

Height — 8 feet. 

Take moments about the outer edge of the base of the pier. 

We shall then have, — 

i°. Moment of vertical load on dome, and of weight of dome 
sector about inner edge of springing, = 

(12.512) (8.68 - 6.56) = 26.52. 

2°. Moment of same about outer edge of springing of pier = 
26.52 + (12.512)/. 



656 APPLIED MECHANICS. 

3 . Moment of horizontal forces about the same axis == 

38.730 + (4.32) (5.41) = 62.101. 

4 . Moment of weight of pier about outer edge = 

\ 8 (1.5 t)t\\t = 6.04/ 2 . 
Hence we have 

6.04/ 2 -f- 12. 51/* + 26.52 = 62.101 

.\ P +. 2.07/ = 5.89 .*. / = 1.60 feet. 

This is the thickness required in order that the line of 
resistance may remain within the lower joint. 

If, on the other hand, while pursuing the same method with 
the dome itself, we require that the line of resistance shall 
remain within the middle third of the pier, we take moments 
about a point in the springing of the pier at a distance \t from 
its inner edge, we should then have 

|* + |(2.o 7 )/= 5-89 

.-. t 2 -f 2.07/ == 8.84 /. / = 2.10 feet. 

On the other hand, we could proceed in a similar way to 
the above, if we desired to keep the line of resistance in the 
dome within the middle third, by merely assuming the horizon- 
tal thrusts to act at two-thirds the thickness of a joint from the 
lower edge, and using a point two-thirds the thickness from 
the top, instead of the lower edge, as the lower limiting-point 
for the pressure to pass through. 

This will not be done here, however. 

Example. — As an example, St. Peter's dome will be given, 
with the dimensions as given by Scheffler reduced to English 
measures. The dome consists in its upper part, as will be 
evident from the figure, of two domes ; the lantern resting on 



DOMES. 



657 



the two is assumed to have one-third of its weight resting on 
the upper, and two-thirds on the 
lower dome. 

Diameter of dome = diameter 
at the base = 144 feet. 

Up to a point 28.48 feet above 
the point C\t is formed of a single 
dome 11.84 f ee t thick. In its 
upper part, on the other hand, 
it is composed of two domes 
whose normal distance apart is 
5.15 feet; the exterior having a 
thickness of. 2.56 feet, and the 
inner of 4.13 feet at the top and 
5.15 feet at the springing. At 
the top of these two domes is an 
opening 12.24 ^ eet radius, sur- 
mounted by a cylindrical lantern. 
The magnitude of the load of the 
lantern on the dome is repre- 
sented on the figure by 1.82 feet 
width and 56.66 feet height. 

Height of the entablature 
ABCD = 23.69. 

Width of ABCD normal to 
plane of paper =: 1.02 feet. 

Thickness of ABCD = 10.30 
feet. 

Divide the exterior dome into 
nine parts, the interior into eight 
of a uniform circumferential width of 10.08 feet, except the 
first, which has a width of only 1.82 feet. 

Determine whether this thickness of ABCD is sufficient to 
keep the line of resistance within joint AB. 




Fig. 295. 



658 APPLIED MECHANICS. 



CHAPTER X. 

THEORY OF ELASTICITY, AND APPLICATIONS. 

§ 275. Strains. — When a body is subjected to the action 
of external forces, and in consequence of this undergoes a 
change of form, it will be found that lines drawn within the 
body are changed, by the action of these external forces, in 
length, in direction, or in both ; and the entire change of form 
of the body may be correctly described by describing a suffi- 
cient number of these changes. 

If we join two points, A and B, of a body before the exter- 
nal forces are applied, and find, that, after the application of 
the external forces, the line joining the same two points of the 
body has undergone a change of length A(AB), then is the limit 

of the ratio , as AB approaches zero, called the strain of 

the body at the point A in the direction AB. 

If AB -f- il(AB) > AB, the strain is one "of tension ; whereas, 
if AB + A(AB) < AB, the strain is one of compression. 

In order to study the changes of form of the body, let us 
assume a point O within the body when there are no external 
forces acting, and let us draw through this point three rectangu- 
lar axes, OX, OY, and OZ, and assume a small rectangular 
parallelopipedical particle whose three edges are OA, OB, and 



STRAINS. 



659 



OC, and let us examine the form of this particle after the 

loads are applied ; it will be 

found that the edges OA, OB, 

and OC will be of different 

lengths from what they were 

before, and that the angles 

A OB, AOC, and BOC will no 

longer be right angles, but 

will differ" slightly from 90 . 

Let the parallelopiped oabc- 

g dcf represent the form and 

dimensions of the particles 

after the external forces are 

applied. Then we shall have, 

if e_„ ej,, and e s represent the strains in the directions OX, O F, 

and OZ respectively, that 

* x — limit of as OA approaches zero, 

c y = limit of as OB approaches zero, 

€ s = limit of — — as OC approaches zero. 




In the figure, e^ and e 2 are tensile strains, and t y is a com- 
pressive strain. 

But these strains do not represent completely the distortion 
of the particle ; for the plane CEGD has slid by the plane 
OABF through the distance cc„ the distance apart of these 
planes being OC, and the plane halfway between the two has slid 
just half as far, so that the amount of shearing, or the shearing- 
strain of planes parallel to XOY in the direction OX, may be 



represented by 



OC 



— - nearly, or the distortion divided by the 
cc x 



660 APPLIED MECHANICS. 

distance apart of these planes. This, moreover, is the tangent 
of the angle occ iy or the tangent of the angle by which aoc 
differs from a right angle. 
If, now, we let 
y zx = shearing-strain in a plane perpendicular to OZ in the 

direction OX, 
y zy = shearing-strain in a plane perpendicular to OZ in the 

direction OY, 
y yx = shearing-strain in a plane perpendicular to O Fin the 

direction OX, 
y yz = shearing-strain in a plane perpendicular to OY in the 

direction OZ y 
y xz = shearing-strain in a plane perpendicular to OX in the 

direction OZ, 
y xy = shearing-strain in a plane perpendicular to OX in the 

direction OY, 

and let boc = <£, aoc = ib, aob = y, then we shall 

2 2 2 

have 

y 2 ^r = — ■ = tan iff, y yz = tan <f>, 

y zy = tan <J>, y xz = tan if/, , 

y yx = tan x , y xy = tanx- 

We thus have 

7^ = yy* = -tan^, 

y^z = y zx = tan i/r, 

7^ = 7yx = tan x, . 

three very important equations. 

We thus have to determine six strains, in order to define 
completely the state of strain in a body at a given point ; viz., 
if we assume three rectangular axes, we must know c x , e y , e z , 
y zy — y yz , Jzx = y*z, Ixy = 7yx> three normal and three tangen- 
tial strains. 



STRAINS IN TERMS OF DISTORTIONS. 66 1 

§ 276. Strains in Terms of Distortions. — Let us assume 
a rectangular parallelopipedical particle, the co-ordinates of one 
corner of which are x, y, z, and of the other, x -f- dx, y + dy, 
z + dz ; this being the case before the load is applied. 

Let the effect of the load be to change x, y, z, respectively, 
into x + £, y -f- rj, z + & and to change x + dx, y + dy, z + 
<fe, into (x + • fl -f (*r> + #), (jf ■ + ,) + (^ + d v ), (z + 
+ (<afe + ^0- Then are dx, dy, dz, the edges of the particle 
before the load is applied. 

Then, from what has preceded, we shall have 



_d£ _d v _d£ 

dx dy dz 



d£ dr, d£ dt d v , d' 

y„ s *y r =- + - s y xz = y zx =- + -, y yz = y zy = - + _ ; . 



The first three will be evident at once. As to the last three, 
if the student will construct the figure indicated, he will see 
that 

— -f -± = tan x , — + — = tantfr, -J- + — = tanc/>. 
dy dx dz dx dz dy 

§ 277. Determination of the Strain in any Given Direc- 
tion. — Suppose we are required, knowing the strains «*, e yj c z , 
y xy , y xz , y yz , to determine the strain in a direction making angles 
a> A y, with OX, O Y, OZ respectively. Assume our rectangu- 
lar parallelopipedical particle in such a way that the diagonal 
from (x, y, z) to (x -f- dx, y -f- dy, z + dz) shall be in the 
required direction, and call the length of this diagonal ds ; then 
we shall have 

(dsy = (dxy + (d y y + {&)*, (i) 

dx , . 

cos a = — , (2) 



662 APPLIED MECHANICS. 

cos /J = * (3) 

dz f \ 

cos r = -. (4) 

Let € be the strain in the required direction ; then length of 
diagonal after load is applied will be 

ds{i + 0, 

and we shall have 

(ds) 2 (i + c) 2 = (dx + d£Y + (dy + drj) 2 + (dz + &)*, 
or 
(ds) 2 + 2e(ds) 2 -f e 2 ^) 2 = (dx) 2 + (dy) 2 + (</s) 2 + 2(<ftrd£ 

+ ^ + dzdO + (d£) 2 + (drj) 2 + (d§ 2 . (5) 

Now, subtracting (i) from (5), and neglecting e 2 (ds) 2 , (d£) 2 , 
(drj) 2 , and (d'£f as being very small compared with the rest, we 
have 

26 (ds) 2 = 2dxd£ + zdyd-q + zdzdt, 



or 
But 



eds = £# + «* + $* (6) 

tfJ <W flTJ 



«ft = d^cosa + drj cos ft + a^cosy. (7) 



# = ^& + f ^ + S*, (8) 

^r </y "2 

A, = ^ + £* + **, (9) 

#x /rv dz 

dl = £& + §dy + §dz. (10) 

<&; dy dz 



srxzssEs. 663 

Hence, substituting these, we have, after dividing by ds, and 
observing (2), (3), and (4), 

d£ , . dn ■> n . dt> „ /d-n , dt\ 

c = — cos 2 a H ! cos 2 B H cos 2 7 -f ( — H )cos B cos 7 

dk ^ dz Vfe ^/ 

/d£ d£\ /dn d£\ 

+ (-f- H- — -)cosacos7-|- (-7- + - 1 ) cos a cos/?/ (11). 
\^s ^a:/ Wx dyj 

or, making use of § 276, we have 

€ = e^C0S 2 a -j- CyCOS 2 /? + e^ cos 2 7 -f- y yz cos ^8 cos 7 

+ 7^2 cos a cos 7 + 7.Z3, cos a cos /?, (12) 

which gives us the strain in any direction. 

It can be shown that there are three directions, at right 
angles to each other, that give the maximum strains or mini- 
mum strains : and we might deduce the ellipsoid of strains, in 
which semi-diameters of the ellipsoid represent the strains ; but 
we will pass on to the consideration of the stresses. 

§ 278. Stresses. — When a body is subjected to the action 
of external forces, if we imagine a plane section dividing the 
body into two parts, the force with which one part of the body 
acts upon the other at this plane is called the stress on the 
plane ; and, in order to know it completely, we must know its 
distribution and its direction at each point of the plane. If we 
consider a small area in this plane, including the point O, and 
represent the stress on this area by/, whereas the area itself is 

P 
represented by a, then will the limit of — , as a approaches zero, 

be the intensity of the stress on the plane under consideration 
at the point O. Observe that we cannot speak of the stress at 
a certain point of a body unless we refer it to a certain plane 
of action : thus, if a body be in a state of strain, we do not 
attempt to analyze all the molecular forces with which any one 



664 



APPLIED MECHANICS. 



particle is acted on by its neighbors : but, when we assume a 
certain plane of section through the point, the stress on this 
plane at the point becomes recognizable in magnitude and 
direction ; and what the magnitude and direction of the stress 
at the given point is, depends upon the direction of the plane 
section chosen, the magnitude and direction differing for differ- 
ent plane sections through the point. 

§279. Simple Stress. — A simple stress is merely a pull 
or a thrust. Assume a prismatic body, with sides parallel to 
OX, subjected to a pull in the direction of its 
length ; the magnitude of the pull being P. As- 
sume first a plane section AA normal to the 
direction of P, and let area of AA be A. Then, 
if p x represent the intensity of stress at any 
point of this plane, 




_ P 

P*~ A 

This, which is the intensity of the stress as dis- 
tributed over a plane normal to its direction, may 
i be called its normal intensity. 

fig. 2 97 . Q n ^ Q^gj. hano^ if we desire to ascertain 

the intensity of the stress on the oblique plane BB, making an 
angle with AA, we shall have 



Area BB = 



cos 



Hence, if p r represent the intensity of the stress on this plane 
in the direction OX, we shall have 



A = 



(— ) 

\cos $S 



— cos V — p x COS V. 
A r 



CO 



If we resolve this into two components, acting respectively nor- 



COMPOUND STRESS. 66$ 

mal and tangential to BB, and if we denote the normal intensity 
by/«, and the tangential by p ti we shall have 

p n = p r cos 6 = p x cos 2 0, (2) 

p t — p r sinO = p x cosOsinO. (3) 

If, now, we assume another oblique plane section, perpen- 
dicular to the first, we shall obtain the normal/,/ and the tan- 
gential pt stress on this plane by substituting for 0, 0; 

2 
hence we obtain 



Hence follows 



p H '=p x sm*6, (4) 

// = p x cos sin 0. (5 ) 



==A; 



or, the tangential components of a simple stress on a pair of 
planes at right angles to each other are equal. 

§280. Compound Stress. — A compound stress may be 
accounted to be the resultant of a set of simple stresses, and 
may be analyzed into different groups of simple stresses. 

Proposition. — Whatever be the external forces applied to a 
body, if through any point we pass three planes of section at right 
angles to each other, the tangential components of the stress on 
any two of these planes in directions parallel to the third must 
be of equal intensity. 

To prove this proposition, assume 
three rectangular axes, origin at O, and 
assume a rectangular parallelopipedical 
particle, as shown in the figure, so 
small that we may without appreciable 
error assume the stress on any one of 
the faces to be the same as that on the 
opposite face ; resolve these stresses, 

i.e., the forces exerted upon the faces of the particle by the 
other parts of the body, into components parallel to the axes. 



r. 


r 


^ 


3 / 





f: 

E 


./ 


A 


/ 
z I 


':/ 


/ / 




Jv 




/a 


G 


/~~ 


> X 


«^, 


x t o, 




/ / 


L A 

% 


B 


1 


V 


/' 


v/< 


i ' 


\ 


I* 


F 



666 APPLIED MECHANICS. 

Let <j x = intensity of normal stress on the x plane, 
a-y — intensity of normal stress on the y plane, 
<t z = intensity of normal stress on the z plane, 
r xy = intensity of shearing-stress on x plane in direction 

OY, 
Txz ±= intensity of shearing-stress on x plane in direction 

OZ, 
Tyjt = intensity of shearing-stress on y plane in direction 

OX, 
Ty Z = intensity of shearing-stress on y plane in direction 

OZ, 
t zx = intensity of shearing-stress on z plane in direction 

OX, 
Tzy ■=. intensity of shearing-stress on z plane in direction 
OY. J 
We have thus apparently nine stresses, which must be given, 
in order to define the stress at the point O completely ; but we 
will now proceed to prove that 

T ' xy == Tyxi Tjcz ~~ ^zxi Tzy ~~ Ty Z . 

In the figure, the only ones of these stresses that are repre- 
sented are, the following : — 

Xa = Xid.! = <j x , 

yP = yfit = vy, 

y/3 2 = yfi z = Tyx , 
Zy = Zf( x = cr z . 

The other four are omitted, in order not to complicate the 
figure. 

Now, it is evident that the total normal force on the face 
AFGD and the normal force on the face OBEC balance each 
other independently, and likewise with the other normal forces. 



GENERAL REMARKS. 66? 

The only forces tending to cause rotation around OZ are 
the equal and opposite parallel forces r xy (area AFGD), one act- 
ing on the face AFGD, and the other on the face OBEC ; and 
the equal and opposite forces r yx (area FBEG), one acting on 
the face FBEG, and the other on the face CO AD. 

The first pair forms a couple whose moment is r xy (area 
AFGD) (xx t ), and the second has the moment r yx (area FBEG) 

But 

Area AFGD = {FA) (zz,), area FBEG = (FB) (zz t ) 

/. r xy (FA) (z Zl ) (xx,) = r yx (FB) (zz x ) (y 7l ) . 

Cancelling zz„ we have 

r xy (FA)(xx I ) = r yx (FB)( yyi ). 



But 



FA — yy I and FB = xx x 
r xy (xx 1 )(yy 1 ) = r yx (xx I )(yy l ) 



Q. E. D. 

In a similar manner we can prove 

T yZ = T^y. 

GENERAL REMARKS. 

From what precedes, it follows, that, when we have the six 
stresses 

°~X) °"yt °*2> T xy , T xz , Tyzi 

or, in other words, the normal and tangential components of 
the stresses on three planes at right angles to each other, given, 
the state of stress at that point is entirely determined ; and, 
when these are given, it is possible to determine the direction 
and intensity of the stress on any given plane. 



66S APPLIED MECHANICS. 

Moreover, if three rectangular axes, OX, OY, and OZ, be 
assumed, and the direct strains along these axes be given, and 
also the shearing-strain about these axes, then the direct strain 
in any given direction can be determined, and also the shearing- 
strain around this direction as an axis. 

The two above-stated propositions furnish two of the funda- 
mental propositions of the theory of elasticity, the third being 
the determination of the relation between the stresses and the 
strains. 

§ 281. Relations Governing the Variation of the Stresses 
at Different Points of a Body. — If we assume a point whose 
co-ordinates are (x, y, z), and a small parallelopipedical particle 
having this point and the point (x -f- dx, y + dy, z + dz) for the 
extremities of its diagonal, we shall have, for the edges of this 
particle, dx, dy, dz, respectively. 

Now let the stresses at (x, y, z) be 

a xi °j/j °"z> r xyy Txzi ^yz y 

i.e, <r x denotes the normal stress on any plane perpendicular to 
OX, and passing through the point (x, y, z), etc. Then, for 
the planes passing through (x + dx, y + dy, z + dz), we shall 
have the stresses 

a x -f d(j x , o-y -f- da-y, a- z -f- dcr z , T xy + dr xy , r xZ -f dr xz , r yz + dr yz . 

We may also have outside forces acting upon the particle in 
question : if such is the case, let the components of the result- 
ant external force along the axes be respectively 

Xdxdydz, Ydxdydz, Zdxdydz. 

Now impose the conditions of equilibrium between all the 
forces acting on the particle. To do this, place equal to zero 
the algebraic sum of all the forces parallel to each of the axes 



RELATIONS BETWEEN STRESSES AND STRAINS. 669 

respectively, the moment equations having already been incor- 
porated in our demonstration that 

T xy = r yxi T xz = T zx> Ty Z — T zy . 

Hence we have three conditions of equilibrium, as follows : — 

(Ox+dox — Ox)dydz + {T xy +dT xy — T xy )dxdz+ [r xz +dT xz — T xz )dxdy+Xdxdydz = o, 
(Oy +da y — c y )dxdz + {r xy -\-dr xy —r xy )dydz +{Ty Z +dTy Z — r yz )dydx+ Ydxdydz = o, 
(a z +da z —,a z )dxdy + {r yz +dr yz — r yz )dxdz+ {r xz +dT X z — r xz )dzdy + Zdxdydz = o. 

Hence, reducing, and dividing by dxdydz, we have 

dcr x dr-rv dr X z 

_£ + __^ + _^ + x= o, (1) 

dx dy dz 

^+^+^ + K=o, (a) 

dx dy dz 

dr xz dr yz dcr z 

-^ + lfy + -^ +z=0 - (3) 

If the particle is in the interior of the body, so that no ex- 
ternal forces act upon it, then X = Y = Z — o. 

Equations (1), (2), and (3) give the necessary relations which 
the variations of stress from point to point must satisfy in order 
that the conditions of equilibrium may be fulfilled. 

§ 282. Relations between the Stresses and Strains. — 
Before proceeding to the general problems of composition of 
stresses, i.e., of determining from a sufficient number of data 
the stress upon any plane, we will first discuss the relations 
between the stresses and the strains ; and we will confine our- 
selves to those bodies that are homogeneous, and of the same 
elasticity throughout. 

From what we have already seen, if to a straight rod whose 
cross-section is A there be applied a pull P in the direction of 



67O APPLIED MECHANICS. 

its length, the intensity of the stress on the cross-section wi 
be 

P 

and, if E be the tensile modulus of elasticity of the material of 
the rod, the strain in a direction at right angles to the cross- 
section, or, in other words, in the direction of the pull, will be 



Now, another fact, which we have thus far taken no account 
of, is, that although there is no stress in a direction at right 
angles to the pull, or, in other words, although a section at 
right angles to the above-stated cross-section will have no stress 
upon it, yet there will be a strain in all directions at right angles 
to the direction of the pull : and this strain will be, for any direc- 
tion at right angles to the pull, 

€ I = — , 

m 

being of the opposite kind from c ; thus, if e is extension, «, is 
compression, and vice versa. 

Hence, if, at any point O of such a rod, we assume three 
rectangular axes, of which OX is in the direction of the pull, 
and we use the notation already adopted, we shall have 



, CTy CTg — T X y — T xz Ty Z O, 



_ a x I &X _ *x 

£ x — , €y € Z — j 

A m A m 

y*y = Jxz = Jyz = o. 



RELATIONS BETWEEN STRESSES AND STRAINS. 67 1 
MODULUS OF SHEARING ELASTICITY. 

In the case of direct tension or compression, when only a 
simple stress is applied, we have defined the modulus of elas- 
ticity as the ratio of the stress to the strain in its own direction. 

Adopting a similar definition in the case of shearing, we 
shall have 

r xy T xz r yz ~ 

l*y I/** lyz 

where G is the modulus of shearing elasticity. 

GENERAL RELATIONS BETWEEN STRESSES AND STRAINS. 

Whenever a compound stress acts on a body at a given 
point, let the stresses be 

°\r> °"y> ^z) T xy> r xz> Ty Z j 

then we shall have, for the strain in the direction OX, 



<*x 


I (Ty 


I <r z 


T xy 

y* y - -£, 


* x E ~ 


m E 


m E' 


^ - °* - 


I <r x 


1 <r z 


Txz 

7- = -g. 


* E 


m E 


m E' 


" = E- 


I <*x _ 

m E 


I CTy 

m E 1 


Ty Z 

yyz - -g' 



This enables us to determine the strains in terms of the 
stresses, as soon as the values of E, G, and m are known from 
experiment, for the material under consideration. 

If, on the other hand, the stresses be required in terms of 
the strains, we can consider e xt e y , <- 2 , y xy , y xza y y2y as known, and 
determine o- X) <r yf <r z , r xy} r yzi r xzt from the above equations. 



6j2 



APPLIED MECHANICS. 



We thus obtain 



E*X = <Tx 

Ety = (Ty 

Ez z = <T Z 



(Ty 


+ 


O"* 




m 




&X 


+ 


^2 




m 




<y.x 


+ 


(Ty 



(I) 

(3) 



and, by solving these equations for the stresses, we have 

(5) 



m 

m -f- 

m 



*X 4- fy + 



m 

in 



-A 

^ E L + * + '> + '•) 

-f- i \ m — 2 / 



(6) 



and also 



= (Ty, 



(7) 



= G : 



(8) r^ = Gy, 2 . (9) 



These equations express the stresses in terms of the strains. 
The three last might be written as follows (see § 276) : — 



\dy dx) 
\dz dx) 



(10) 
(") 

(12) 



as these forms are often convenient. 

§ 283. Case when <r g = o. — Inasmuch as there are many 
cases in practice where the stress is all parallel to one plane, 
and where, consequently, the stress on any plane parallel to 
this plane has no normal component, it will be convenient to 
have the reduced forms of equations (4), (5), and (6) which 
apply in this case. 



VALUES OF E, G, AND m. 673 

Let the plane to which the stresses are parallel be the Z 
plane ; then a z = o. Then equation (6) becomes 

. *x 4- €y + ** 

«* H = o 

m — 2 

«r + €y 



€ 2 = — 



(#* — 2)(m — 1 ) 



and, substituting this value of e 2 in (4) and (5), and reducing, we 
obtain 



T x = ; E\ € x H h (i) 

m + 1 \ m — 1} w 

>> = — r— 4^ + - — > o) 



which are the required forms. 

The other three equations, viz., 



= £y.rp T*s = Gy*s, Tyz = &y 



yzt 



remain the same as before. 

§284. Values of E, G, and m These three constants 

need to be known, to use the relations developed above. 

i°. As to E, this is the modulus of elasticity for tension, 
and has been determined experimentally for the various mate- 
rials, as has been already explained. Moreover, it has also been 
shown experimentally, that, with moderate loads, the modulus 
of elasticity for compression is nearly identical with that for 
tension in cast-iron, wrought-iron, and steel. 

2°. As to m, in those few applications that Professor Ran- 
kine gives of his theory of internal stress, such as the case of 
combined twisting and bending, he determines the greatest in- 
tensity of the stress acting ; and his criterion is, that this shall 
be kept within the working-strength of the material. This is 
equivalent to assuming m = 00. The more modern writers, 



674 APPLIED MECHANICS. 

such as Grashof and others, take account of the fact that m has 
a finite value, and make their criterion that the greatest strain 
shall be kept within the quotient obtained by dividing the work- 
ing-strength by the modulus of elasticity of the material. 

Thus, if / is the working-strength, and <r I the greatest 
stress, and « r the greatest strain, Rankine's criterion of safety 
is 

whereas the more modern criterion is 

The resulting formulae differ in each case ; and, as has been 
stated, those of Rankine could be derived from the more gen- 
eral ones by making 

i 

~ — o or m = co, 

which is never the case. 

As to the value of m, but few experiments have been made. 
Those of Wertheim give, for brass, 2.94 ; for wrought-iron, 3.64. 

The values m = 3 and m = 4 are those most commonly 
adopted, so that 

11 11 

— — - or — — ~- 
m 3 m 4 

3 . The value of G, the shearing-modulus of elasticity, i.e., 
the ratio of the stress to the strain for shearing, has been 
determined experimentally, and has generally been found to be 
about two-fifths that for tension. 

According to the theory of elasticity, we must have 

G = I -^-E f 
2 m + 1 

as may be proved as follows : — 



VALUES OF E, G, AND m. 675 

Assume a square particle whose side is a, and let a simple 
normal stress o- be applied at the face AB ; then we 
shall have, on the planes BD and AC, a shearing-stress 
(§ 279) 



t = o- sin 45° cos 45° = -Jo*. 




On the other hand, if we let d i c 

r 

cr I 

€ ~p* Fig. 299. 



the strain of the particle in the direction AD will be e, while 

that in the direction AB will be ; hence the particle will 

m 

become a rectangle, the side AD changing its length from a to 
a + tfe, and side AB changing from a to a . 

171 

The diagonals will no longer be at right angles to each 
other ; and, if we denote by a the angle by which their angle 
differs from a right angle, we shall have, for the shearing-strain 
on the planes AC and BD, 

y = tana. 

But, after the distortion, the angle ADB will become 



!(H 



a (Ze e 

tan- a 1 

m m 

tan 



(7T a\ 2 
J = = - 
42/ . , a a 
^ ' 1 -h tan - 



-f- at 1 +e 



therefore, dividing, and carrying the division only to terms of 
the first degree, we have 



a 
2 tan - = 1 



hi} 

(a\ m -f- I 
) = e. 
2/ 7n 



6y6 APPLIED MECHANICS. 



But 



y = tan a = 2 tan - nearly 

2 



7 = 


m -f- i 
m 




T 


¥ 


i w 


y 


m 4- i 

! € 

m 


2 W + I 


T 


G and 


2 = ^ 


7 




€ 


G = 


1 * is. 





but 



2 W + I 

§ 285. Conjugate Stresses. — If the stress on a given plane 
at a given point of a body be in a given direction, the stress at 
the same point on a plane parallel to that direction will be 
parallel to the given plane. Let YO Y represent, in section, a 
given plane, and let the stress on that plane be in the direction 

xox. 

Consider a small prism ABCD within a body, the sides of 
whose base are parallel respectively to XOX and YO Y The 
forces on the plane AB are counterbalanced by the forces 
on the plane DC ; the resultants of each of these sets being 
equal and opposite, and acting along a line passing through 0. 
Hence the forces acting on the planes AD and BC must be bal- 
anced entirely independently of any of the forces on AB or 
DC: and this can be the case only when their, direction is paral- 
lel to YOY ; for otherwise their resultants, though equal in 
magnitude and opposite in direction, would not be directly 
opposite, but would form a couple, and, as there is no equal and 
opposite couple furnished by the forces on the other faces, equi- 
librium could not exist under this supposition. 

§ 286. Composition of Stresses. — The general problem of 
the composition of stresses may be stated as follows : — 



PROBLEM. 



677 



Knowing the stresses at a given point of a strained body 
on three planes passing through that point, to find the stress at 
the same point on any other plane, also passing through the 
same point. The stresses on the three given planes are not 
entirely independent ; in other words, we could not give the 
stresses on these three planes, in magnitude and direction, at 
random, and expect to find the problem a possible one. Thus, 
suppose that the planes are at right angles to each other, we 
have already seen that we have the right to give their three 
normal components, <r x , <r y , and a z , and the three tangential, 
r xy , T yZ y and t xz1 and that T yx =■ r xy , etc. We will now proceed 
to special cases. 

§ 287. Problem. — Given the three planes of action of the 
stress as the x, y, and z plane respectively, and given the nor- 
mal and tangential components of the stresses on these planes, 
viz., <r x , <r y , <r„ r xy , t xz , and r yZf to find the intensity and direction 
of the stress on a plane whose normal makes with OX, OY, 
and OZ the angles a, /?, and y respectively, where, of course, 

COS 2 a + COS 2 /? + C0S 2 y = I. 

Draw the line ON> making angles a, /?, and y with OX, OY, 
and OZ respectively ; then draw 
near O the plane ABC perpen- 
dicular to ON. It has the direc- 
tion of the required plane, and 
cuts off intercepts OA, OB, and 
OC on the axes ; and, moreover, 
we shall have, from trigonometry, 
the relations, 

Area BOC = {ABC) cos a, 

Area AOC = (ABC) cos ft 

Area A OB = (ABC) cosy. 




Fig. 300. 



Now consider the conditions of equilibrium of the tetrahe- 
dron OABC. The stress on ABC must be equal and directly 



6y8 APPLIED MECHANICS. 

opposed to the resultant of the stresses on the three faces 
AOC, BOC, and AOB. Now let us proceed to find this result- 
ant. 

In the direction OX we have the force 

<r x (BOC) + r x% {AOB) + r xy (AOC) 

= (ABC)(cr x cos a -f r xy cos j3 + r^cosy). 

Lay off <9Z> to represent this quantity. In the same way repre- 
sent the force in the direction OY by 

OE = <r y (AOC) 4- r yz {BOA) 4- r xy {BOC) 

= {ABC) (fj y cos /? 4- T xy cos a + Ty 3 cos y) , 

and that in the direction OZ by 

OF = <r z (AOB) 4- r^BOC) 4- r xz (AOC) 

— ABC(cr z cosy + Tj, z COSa -f- t^ 2 cos/5). 

Now compound these three forces, and we have, as resultant 

force, 

F = <9c7 = VOP 2 4- 0£ 2 + OF 2 , 

and as resultant intensity 

R <J0D 2 + 6>^ 2 -f OF 2 



=b \^(<j>cosa + TxyCOs/3 4- T^ 2 cosy) 2 

+ (<Ty COS P + T^COSa + T^COSy) 2 

4- (o- cosy 4- TysCOSa + t^COs/?) 2 ^ 
= Vjo-^ 2 cos 2 a 4- cr y 2 cos 2 ft 4- <r/cos 2 y 

4- r x /(cos 2 a 4- cos 2 /?) 4- r jr2 2 (cos 2 a 4- cos 2 y) 

4- Tyz 2 (cos 2 /? 4- cos 2 y) + 20-^(7-^ cos/? 4- t^ cos y) cos a 

4- 2CTy(T^ >/ COSa 4- Ty' z cos y) COS (3 
4" 2(T 2 (T_j, 2 C0Sa -f T^COS/?) -f 2T^ 2 COS /? COS y 

4- 2 TjcyTyg COS a COS y 4" StptfTav cos a cos/? J; 



STRESSES PARALLEL TO A PLANE. 



679 



the direction being given by the angles, a r , j3 r , and y ry where 

OF , 



COS a r 



OD 
R' 



a 0E 

cos ft. = — , 

K 



cos y r = 



R 



§ 288. Stresses Parallel to a Plane. — To solve the same 
problem when there is no stress in the direction OZ> and when 
the new plane is perpendicular to XOY, or, in other words, in 
the case when the planes of action are all perpendicular to one 
plane, to which the stresses are all parallel : we then have 

0* S53 t xz = Tyz = o and fi — 90 — a, 
and hence 

cr = \cr x 2 COS 2 a -f" <T y 2 sin 2 a -f- r xy 2 + 2 (<r x •+■ <r y )r xy COS a sin a. 

Or we may proceed as follows : — 

Let the normal intensity of the stress on the x plane (i.e., 
that perpendicular to OX) be o- x , that on the 
y plane a- y , and the tangential intensity r xy . 
Let ON be the direction of the normal to 
the plane on which the stress is to be deter- 
mined, and let the angle XON = a. Then 
let the plane AB be drawn perpendicular to 
ON, and let us consider the equilibrium of 
the forces exerted by the other parts of the 
body upon the triangular prism whose base is ABO and alti- 
tude unity. 

If we compound the forces acting on the faces AO and OB, 
we shall have, in their resultant, the total force on the face AB 
in magnitude and direction. Moreover, we have the relations, 

Area OB = area AB cos a and Area OA = area AB sin a. 
Force acting on OB in direction OX = <r x (OB), 
Force acting on OB in direction OY = r xy {OB), 
Force acting on OA in direction OX = r xy {OA), 
Force acting on OA in direction O Y = a- y (OA). 




Fig. 301. 



68o 



APPLIED MECHANICS. 



Hence, if we lay off 

OD = <j x (OB) + r xy (OA) and OC = o-y(OA) + r xy (OB), 

then will 6>Z> represent the total force acting in the direction 
OX, and 6^(7 will represent the total force acting in the direc- 
tion OY. 

Compounding these, we shall have OE as the resultant total 

OF 
force on the face AB, and — — will represent its intensity. 

To deduce the analytical values, we have 

OD = <t x (OB) + r xy (OA) = (AB)(<r x cosa + r^sina), 
OC — (Ty(OA) -h r xy (OB) = (AB) (a y sin a -f r^cosa) 



\ OE = SJO&+OC 2 



— ABy((T x cos a + r^sina) 2 -f- (cr^sina -f- r^cosa) 2 

= AB\\a- x 2 cos 2 a -f ay 2 sin 2 a + 27-.^ cos a sin a.(<r x •+- 0^,) 

+ r X y 2 (cos 2 a 4- sin 2 a) J. 

Or, if oy represent the resultant intensity on the plane AB, and 
a r the angle this resultant makes with OX, we shall have 

o> = \\o- x 2 cos 2 a + o-y 2 sin 2 a 

+ 27-^(0-^ -h o>) COS a sin a -T- t x /\, (i) 



and 



<9Z) , OC 

cosa r ess -— and sin<v = — — . 

0£ 6>^ 



Moreover, it is sometimes desirable to resolve the stress into 
normal and tangential components. If this be done, and if o- 
and t represent respectively the normal and tangential com- 
ponents, we shall have 



OF , EF 

* = AB ^ T = AB> 



PRINCIPAL STRESSES. 68 1 

but 

OF = OD cos a -f- ED sin a and iii 7 = ED cos a — OD sin a 

0Z? , OC . 

cr = COS a + Sin a 

AB AB 

■ = <r*C0S 2 a + <Tj,sin 2 a -|- 2r^ COS a sin a (2) 
and 

OC OD . 

t = COS a — — - sin a 

AB AB 

= (cry — o-^) cos a sin a -f- r^(cos 2 a — sin 2 a) 

(a> — o-^N 
J Sin 2a + T X y COS 2a. (3) 

§ 289. Principal Stresses. — It will next be shown, that, 
whatever be the state of stress in a body, provided the stresses 
are all parallel to one plane, the planes of action being all taken 
perpendicular to this plane, there are always two planes, at right 
angles to each other, on which there is no tangential stress; 
these two planes being called the planes of principal stress, the 
stress on one of these planes being greater, and the other less, 
than that on any other plane through the same point. 

To prove the above, it will be necessary only in the last 
case, which is a perfectly general one, to determine for what 
values of a the value of r is zero, and whether these values of a 
are always possible. We have 

<Ty — cr x # 

t = Sin 2a + r xy COS 2a : 

and, if we put this equal to zero, we have 
sin 2a 2r rv 



tan 2 a 



COS 2a (T x — <J y 



and this gives us, for all values of <r x , cr yy and r xy , two possible 
values for 2a, differing from each other by 180 , hence two 
values for a differing by 90 . Hence follows the first part of 
the proposition. 



682 APPLIED MECHANICS. 

The latter part — that these are the planes of the greatest 
and least stresses — will be shown by differentiating the value 
of o> 2 , and putting the first differential co-efficient equal to zero ; 
and, as this gives us 

2 v x 2 cos a sin a + 2 (Ty 2 cos a sin a 

+ 2T xy (<r x -f <r>,)(cos 2 a — sin 2 a) 

= 2 (cr x + cr y ) \ ((T y — a x ) COS a sin a -f- t^(cos 2 a — sin 2 a) \ 

therefore we have the same condition for the maximum and 
minimum stresses as we have for the planes of no tangential 
stress. 

It follows that the determination of the greatest and least 
stresses at any one point of a body is identical with the deter- 
mination of the principal stresses ; and it will be necessary, 
whenever the stresses on any two planes are given, to be able 
to determine the principal stresses, as one of these is the 
greatest stress at that point of the body, and the other the 
least. 

§ 290. Determination of Principal Stresses When the 

stress is all parallel to one plane, viz., the z plane, and when 
the stresses on two planes at right angles to each other are 
given, i.e., their normal and tangential components, we may be 
required to determine the principal stresses. Proceed as fol- 
lows : Given normal stresses on X and Y planes respectively, 
a- x and o-y, and tangential stress on each plane r xyy to find prin- 
cipal stresses. 

From § 288 we have, for a plane whose normal makes an 
£ngle a with OX, 

°> = V(7> 2 C0S 2 a + cry 2 sin 2 a + 2T X y(cr x -f (Ty) COS a sin a + T X y 2 , (1) 
cr n = <r x COS 2 a -f o> sin 2 a 4" 2T xy cos a sin a, (2) 

r = r^(CQS 2 a — sin 2 a) — (cr x — ay) COS a sin a, (3) 



DETERMINATION OF PRINCIPAL STRESSES. 683 

or 

0"x — °"y 
T = T X y COS 2a Sin 2a. (4) 

Now, the condition that the plane shall be a plane of prin- 
cipal stress is, that t ■= o. Hence write 



xy 



(cos 2 a — sin 2 a) — (a- x — (Ty) cos a sin a = o, 



find a, and substitute its value in (2), and we shall have the 
principal stresses. The operation may be performed as fol- 
lows ; viz., — 

From (3) we have 



1 ( <r x — <r y . ) 

- \ 1 H cos a sin a > . 

2( r xy ) 

I ( o-x — o-y ) 

in 2 a = - { 1 cos a sin a > . 

2 ( T X y ) 

<T X -f- (Ty COS a Sin a ( <T X — 2(T x (Ty + (Ty 2 ) 

"' = ~r~ + — —\ 5 + 4 M 

or 

<r* 4- o> COS a sin a , 1 

<r„ = 1 (o> — o>) 2 4- 41*/ . 



(a) 


"Vr — 0> 


T-ry 




.*. COS 2 a = 


W 


c^ — o> 


T.sry 




.*. sin 2 a = 



Hence 



2T. 



*y 



But we have, since (4) equals zero, 



tan 2a = 



2r 



xy 



.*. Sin 2a = 2 Sin a COS a = 



0* — °> 

±2T X y 



V(o> - cry) 2 4- 4^/ 



684 APPLIED MECHANICS. 

Hence substitute for cos a sin a its value, and 

°» = ^~^ ± ^(**-°>> a +4*V, (5) 

which gives us the magnitudes of the principal stresses ; the 
plus sign corresponding to the greater, and the minus sign to 
the less. 

EXAMPLES. 

i. Let, in the last section, <r y = o, and find the principal stresses. 
Here we have 

2T xy 

tan 2a = 

and 

&x 1 i 

<r n = — ± -\a- x 2 -f 4r 



2 

xy • 



2. Given two principal stresses, to find the stress on a plane whose 
normal makes an angle a with OX. 

In this case r xy = o. 

Hence we have the case of § 288, with the reduction of making 
r xy — o. We may therefore obtain the result by substitution in the 
results of § 288, or we may proceed as follows : — 

(a) Find stress on new plane in direction OX; this will be, § 279, 

cr x COS a. 

{&) Find stress on new plane in direction OY ; this will be, § 279, 

cr y sin a. 
(V) Compound the two, and the resultant is 



o\ 



iy = \l<r x z COS 2 a + a/ sin 2 a. ( I ) 

(d) Normal component of <r x cos a is 

(Ty cos 2 a. 

(e) Normal component of <r y sin a is 

a y sin 2 a. 



ELLIPSE OF STRESS. 



685 



(*) 



(/) Add, and we have, for normal stress, 

a- n = <j x cos 2 a -\- cr y sin 2 a. 

(g) Tangential component of <t x cos a is 

— cr x cos a sin a. 

(h) Tangential component of <r y sin a is 

-\-<j y cos a sin a. 

(£) Add, and we have, for tangential stress, 

t = (<Ty — (T^) cos a sin a. (3) 

§291. Ellipse of Stress. — In the case above, i.e., when 
the two principal stresses are v x and <r y respectively, if we 
represent them graphically by OA = 
a- x and OB = a- p and let CD be the 
plane on which the stress is required, 
its normal making with OX the angle 
XON == a, then, from what has been 
shown, if OR represent the intensity of 
the resultant stress on this plane, we 
shall have 




OR = oy = >Ja- x 2 COS 2 a -f- <t/ sin 2 a ; 

and, moreover, 

OR — a- x cos a, (97^ = <r y sin a. 

If we denote these by x and y respectively, letting (x, y) be 
the point R, i.e., the extremity of the line representing the 
stress on AB, then 



X = oleosa, 



jy = <r y sin a, 



— ) = cos 2 a and ( — ) = sin 2 a 
°V W/ 



jc 2 j' 2 



686 APPLIED MECHANICS. 

which is the equation of an ellipse whose semi-axes are v x and 
a-y respectively ; hence the stress on any plane will be repre- 
sented by some semi-diameter of the ellipse. 



SPECIAL CASES. 

I. When the two given stresses are equal, or c x = <r y} then 



o> = ^a x COS 2 a + a-y 2 sin 2 a = <t x , 

and 

a x COS a 
cos or = = cos a and sin ar = sin a / 



therefore the stress is of the same intensity on all planes, and 
always normal to the plane. 

II. When the two given stresses are equal in magnitude 
but opposite in sign, or <j y = — o- X) then 

o> = a- x . 
But 

cosa r = cos a and sina r = —sin a, 
hence 

a r = — a; 

therefore the stress on any plane whose normal makes an angle 
a with OX is of the same intensity <r x , but makes an angle 
equal to a with OX on the side opposite to that of the normal 
to the plane. 

Problem. — A pair of principal stresses being given, to 
find the positions of the planes on which the shear is greatest. 

Solution. — Let r = (a-y — <r x ) sin a cos a == max. 

Therefore differentiate, and t 

cos 2 a — sin 2 a = o 
.*. cos a = ±sina .*. a = 45 or 135 . 



SPECIAL MODES OF SOLUTION OF SOME PROBLEMS. 687 

§ 292. Some Special Modes of Solution of some Prob- 
lems. — The case where two principal stresses, <j x and a y , are 
given, to find the stress on any plane whose normal makes an 
angle a with OX, may be solved as follows, graphically : — 

Let, Fig. 302, o> = OA, and <r y — OB. Let XON — a. 

Now, 



o> 


+ 


<r x 


+ 


°> 


— 


<r* 




2 






2 


? 


0> 


+ 


&x 




o> 


— 


<?x 



Hence, instead of proceeding at once to find the resultant 

stress on CD due to the action of a x and o- y , we may first find 

that due to the action of the two equal principal stresses of the 

same kind, 

y y + a- x 



then that due to the pair 



and 



2 2 

and then the resultant of these two resultants. 

The first resultant will be evidently laid off on ON, and 

equal in magnitude to — ; hence let OM == y x , and 

OMwiM be the first resultant. 

The second resultant will be of magnitude ~ -; and 

will h^ave a direction MR such that the angle NMS s= SMR. 

Hence, laying off this angle, and making MR =2 — -, 

2 

we shall have for the final resultant, OR, as before. 

This construction will be useful in the following case: — 
To find the most oblique stress, we must find for what 

value of a the angle MOR is greatest. This will be made 



688 APPLIED MECHANICS. 

evident if we observe, that, for all positions of the plane, the 
triangle OMR has always OM = ^ ^ , and MR == ° y ~ ^ 

both of constant length. Hence, if, with M as a centre and 
MR as a radius, a circle were described, and a tangent were 
drawn from to this circle, the point of tangency being taken 
for R, then will OR be the most oblique stress ; i.e., the 
stress is most oblique when ORM = 90 . Therefore greatest 
obliquity = 



sin- 



try 4- o-j 



§ 293. Converse of the Ellipse of Stress. — The converse 
of the ellipse of stress would be the following problem : Given 
any two planes passing through the point in question ; given 
the intensities and directions of the stresses on these planes, 
— to find the principal stresses in magnitude and in direction. 

The first step to be taken is, to assure ourselves that the 
conditions are not incompatible, as they are liable to be if the 
planes and stresses are taken at random. The test . of this 
question is, to resolve each stress into two components, respec- 
tively parallel to the two planes ; and, if the conditions are not 
inconsistent, the component of each stress along the plane on 
which it acts must be equal. The proof of this statement can 
be made in a similar way to that used in proving that the 
intensities of the shearing-stresses on two planes at right angles 
to each other are equal. If, upon applying this test, we find 
that the conditions are not inconsistent, we may proceed as 
follows : — 

Suppose CD (Fig. 302) were the given plane, and OR the 
stress upon it, and suppose the position of the principal axes, 
OX and OY, and, indeed, all the rest of the figure, were absent, 
i.e., not known. Now, we can easily draw the normal ON; 
and, if we could determine upon it the point M such that OM 



CONVERSE OF THE ELLIPSE OF STRESS. 689 

should be one-half the sum of the principal stresses, we should 
be able to reproduce the whole figure. Hence we will devote 
ourselves to the determination of the position of the point M. 

Let OB = p = stress on plane CD. 

Let stress on the other given plane be /,. 

Let NOR = = obliquity of /. 

Let X = obliquity of /,. 

Then we have 

MR 2 == OR 2 -f OM 2 - 2OM. OR cos0 ; 

or, if <r x and o> denote the (unknown) magnitudes of the prin- 
cipal stresses, 

■ 

(^ 2 = ^ + (^>) I - 2 (^>) / cos e . (X) 

From the triangle constructed in the same way, with the 
stress on the other plane, we should have 

Hence, by subtraction, 

^ - p/ = 2 ^_f_ilf> Wcos - p s cos ft) (3) 

O* + 0> /" ~ p? 



2 (p cos 6 — pj. cos Z ) 



(4) 



Having thus found — -, we can next find, from either 

(1) or (2), the value of 



2 

Now, therefore, we know OM and MR, and hence we can lay 
off this value of OM, and complete the triangle OMR ; then 



69O APPLIED MECHANICS. 

bisect the angle NMR, and the line MS is parallel to the axis 
of greater principal stress. Hence draw Y parallel to MS, 
and OX perpendicular to OY, and lay off on Y 



and on OX 



OB = 


o> = 


OM 4- MR 

2 


OA = 


(T X = 


OM -MR 

2 



and the problem is solved. 

§ 294. Case of any Stresses in Space. -r- In the case of 
stress which is not all parallel to one plane, we should find that 
it is always possible,- no matter how complicated the state of 
stress in a body, to find three planes at right angles- to each 
other on which the stress is wholly normal, these being the 
principal stresses ; and a number of propositions follow analo- 
gous to those for stresses all parallel to one plane. The discus- 
sions of these cases become very complex, and will not be 
treated here. .> 

§295. Some Applications. — The following are some of 
the practical cases which require the theory of elasticity for 
their solution. 

§296. Combined Twisting and Bending. — This is the 
case very generally in shafting, as the twist. is necessary for 
the transmission of power, and the bending is due to the weight 
of the pulleys and shafting, and the pull of the belts, this being 
especially so when there are pulleys elsewhere than close to the 
hangers ; also in overhanging shafts, in crank-shafts, etc. 

Thus far we have no tests of - shafting under combined 
twisting and bending, arid" therefore the methods used' for 
calculating such shafts vary. With many it is the practice 
to compute their proper size from the twisting-moment only, 
but to make up for the bending by using a large factor of 
safety, the magnitude of this factor depending upon how much 



COMBINED TWISTING AND BENDING. 69 1 

the computer imagines the shaft will be weakened by the par- 
ticular bending to which it is subjected. 

With others it is customary to compute the deflections, 
under the greatest belt-pulls that can come upon it, by the 
principles of transverse stress, without any reference to the 
torsion, and to so determine it that the deflection computed in 
this way should not exceed -^§-§ or ^oo °^ tne s P an - 

On the other hand, Unwin and some others give the for- 
mulae, which will be developed here for combined twisting and 
bending, as deduced by the theory of elasticity. This formula 
has not, as yet, been very extensively used ; and its constants 
are taken from experiments on tension or torsion alone, and 
not on a combination of the two. It is to be hoped that we 
may some time have some experiments on such a combination. 
We will now proceed to deduce a formula for the greatest in- 
tensity of the stress at any point of the shaft. 

For this purpose 

Let M l b= bending-moment at any section. 

M 2 = twisting-moment at the same section. 

I x == moment of inertia about neutral axis for bending. 

T 2 =. moment of inertia about axis of shaft. 

r = distance from axis to outside fibre. 

Then, if we denote by <x the greatest intensity of the stress 
due to bending, r and by t the greatest intensity of the stress due 
to twisting, we have, 

M\r ~M 2 r 

O- = — . (i) T = — . (2) 

-M ._, ■ . ■** 

For a circular or hollow circular shaft, 

I 2 = 2l^ 



hence 



°- = ~r- (3) t = -— . (4) 



6g2 APPLIED MECHANICS. 

Then, at a point at the outside of the shaft in the section 
under consideration, we shall have, — 
i°. On a plane normal to the axis, 

(a) a normal stress <r, 

(b) a shearing-stress t. 

2°. On a plane in the direction of the axis, 

(a) a normal stress o-. 

(b) a shearing-stress t. 

We thus have the case solved in Example I., § 290. 
If, therefore, the greatest and least principal stresses be 
denoted by <j 1 and <r 2 respectively, we shall have 



ri = ->v/f^, (5) 



%r*. (6) 

2 ▼ 4 

But, if e x and e 2 denote the strains in the directions of the 
principal stresses, we have 

77 °" 2 77 °" 1 ' 

m m 

Hence, substituting for <r l and o- 2 their values, we have 

„ m — 1 ;« + 1 1 , v 

JE Cl = o- + y^ + 4T 2 , (7) 

2*tf 2W 

m — 1 w -f- 1 



^ 2 = -<r - =^-V„. + 4T ». (8) 

2« 2#Z 

We then have, for the greatest stress on any fibre, the 
greater of the two quantities (7) and (8); and this should not at 
any section of the shaft exceed the working-strength of the 
material for tension. 



COMBINED TWISTING AND BENDING. 693 

The greater of the two is 2se x : hence we should have, if 
/ = greatest stress, 

m — 1 . m 4- 1 



<r + ---V^T^=/ (9) 



2W 2W 



If, now, we let m = 4, as is commonly done, we have 



frr + $V'o a + 4T a =/, (10) 

this being the formula given by Grashof and others for com- 
bined twisting and bending. 

On the other hand, Rankine puts the value of <r t in (5) equal 
to /, and hence Rankine's formula is 



<r 1 



2 + -VW4T a =/ (11) 

This might be derived from (9) by making m ^= 00 instead 
of m = 4. 

The formulae developed above are applicable to any section. 



APPLICATION TO CIRCULAR AND HOLLOW CIRCULAR SHAFTS. 

Substituting for o- and t in (10) the values from (3) and (4), 
we should obtain 



jfyft + ^IWTM}\ =/, (12) 



which is Grashof s formula, and is given by Unwin and others ; 
and, substituting in (11) instead, we should have 



-^(M x + )/M> + M 2 >) =/. (13) 

Equation (12) is equivalent to the following rule : — 



694 APPLIED MECHANICS. 

Calculate the shaft as though it were subjected to a b ending- 
moment . . 

M = \M t + \>lMr+W\ ' 

and equation (13) is equivalent to the following rule: — 

Calculate the shaft as though it were subjected to a bending- 
moment 



M 



M l 1 , . 



Now, if, as is usually the case, the section where the great- 
est bending-moment acts is also subjected to the greatest 
twisting-moment, it will only be necessary to put for M, the 
greatest bending-moment, and for M 2 the greatest twisting- 
moment. 

§ 297. Thick Hollow Cylinders subjected to a Uniform 
Normal Pressure. — Let inside radius = r, outside radius = 
r x , length of portion under consideration = unity, intensity of 
internal normal pressure == P, of external normal pressure 

= P v 

i°. Divide the cylinder into a series of concentric rings ; 
let radius of any ring be p, and thickness dp, these being the 
dimensions before the pressure is applied. 

Let p become p -f- £, and dp, dp + dg, after the pressure is 
applied. 

Then at any point of this ring we shall have, for the strain 
in the radial direction, 

t- : 

and, since the length of the ring before the application of the 
pressure is 2?rr, and after is 2ir(r + g), hence the strain in a 
direction at right angles to the radius is 

5=* = i (2) 

2rrp p 



HOLLOW CYLINDERS SUBJECTED TO PRESSURE. 695 

2°. Impose, now, the conditions of equilibrium upon the 
forces exerted by the rest of the cylinder upon the .upper, half- 
ring. For this purpose let 

p = intensity of normal pressure on inside; i.e., at dis- 

tance p from the axis. 
p + dp = intensity of normal pressure on outside ; i.e., at dis- 
tance p + dp from the axis. 
Then we shall have for these forces, — - 

(a) Upward force due to internal pressure, 

a/G> + £). '"' 

(b) Downward force due to external pressure, 

f *(J + d f)i? + £ + d ? + O- 

(c) Upward force at right angles to radius acting at division 
line between the two half-rings, , 

2i(dp + d£), . 

where t = intensity of hoop-tension per square unit ; i.e., of 
tension in a circumferential direction. Then we have 

2 (J + dp)( P -f £ + d P + di) - 2 p( P + i) - 2t{dp + dS) = o; 

and, if this be reduced, and the terms 

2pd£, 2$dp, 2 dp dp t 2dpd%, and 2td£ 

be omitted, all of which are very small compared with the 
remaining ones, we shall have 

d P + — = °' (3) 



Now, the two stresses / and t are principal stresses, since 



696 APPLIED MECHANICS. 

there are no shearing-stresses on these planes. Hence we have, 
from equations (1) and (2), §282, 

(4) 
(5) 

Now eliminate / and / between (3), (4), and (5), and obtain 
a differential equation between p and £■ -■ 
Proceed as follows : — 
From (4) and (5), 



E0- 
dp 


= /- 


m 


d -. 

P 


= / - 


m 



m 2 — i\dp mp] 



dp m 
From (4) and (5) also, 



dp Em 2 /d 2 $ 1 dg £ \ 

— i\dp 2 mp dp mp 2 ) 



p — t Em Idg A 
p ~ m -f i\dp p) 



idp 
Hence, substituting in (3), and reducing, we obtain 





d!l + idi 
dp 2 p dp 


P 2 


= 







dp 2 ' 


=-(-; 


dp 


¥ : 


dp 


Hence, 


by integration, 














d$ 
dp 


-- + 2a; 
p 





(6) 



(7) 



2a being an arbitrary constant, to be determined from the con- 
ditions of the problem. 



HOLLOW CYLINDERS SUBJECTED TO PRESSURE. 6g, 



From (7) we obtain 

^ + *-*V or -^- 



Hence, integrating, we have 

fp = tfp » + £ (8) 

£ being another arbitrary constant. 
From (8) we obtain 

£= *p + -, (9) 

P 

which gives us, for the two strains, 

T= a -- Z > ( I0 ) 

- = *+ ? (11) 

Hence, substituting these values in (4) and (5), and solving 
for/ and t successively, we obtain 



Em Em b 

p = a , 

m — 1 m -f- 1 p 2 


(12) 


. Em . Em b 
t = a -\ . 


(13) 



m — 1 m + 1 p 2 

Now, to determine a and £, we have the conditions, that, 
when p = r, p = P, and when p — r lt p = —P x . 
Hence 

D i?/tf Em b D Em Em b 

1^ = a , .Ti = a — 



m — 1 ;// + 1 r 2 w — 1 m -h 1 rf 

. „ w - 1 P x r 2 — Pr 2 .... m + x p t — p 

.-. # = — 1 —± — , b = • r 2 r L 2 , 

Em r x 2 — r 2 m r? — r 2 

,j t = P z r 2 - Pr 2 1 (/> - P)r 2 r^ , , 

r 2 — r 2 p 2 r 2 — r 2 ' K ' 



698 



APPLIED MECHANICS. 



The greatest value of /, and hence the greatest intensity of 
the hoop-tension, occurs when P == r; and hence we obtain 



Max/ 



2P 1 r l 2 — P(r* + r 2 ) 

r 2 _ rZ 



(15) 



this value of / being negative when there is hoop-tension, be- 
cause the signs were so chosen as to make t positive when 
denoting compression. 

If P s = o, i.e., if there is no external pressure, we have 



Max/ 



■o/ r* + r 2 \ 
\rS ■- r 2 ) 



(16) 



and, according to Professor Rankine's method, we should deter- 
mine the proper dimensions by keeping max / within the work- 
ing-strength of the material. _ f 

On the other hand, if we decide that we will keep the value 

of £[-) within the working-strength, we shall find for this, when 

we make p = r, 

SjzP.r 2 — P(r 2 + r 2 )] - -P(r* — r 2 ) 

; (17) 



and, if m == 4, 

Maxi$ = 
When P, = o, 



(r* _ r 2 ) 

2P,r 2 — P(^ 2 4- ^ 2 ) - P*W - r 2 ) 
<V X 2 - r 2 ) ! 



(18) 



MaX ^U = rf-.rl ' _ (I9) 



Practical cases of thick, hollow cylinders subjected to a uni- 
form normal pressure occur in hydraulic presses and in ord- 
nance. 



STRENGTH OF FLAT PLATES. 



699 



§ 298. Strength of Flat Plates. — In this regard, the for- 
mulae that will be deduced are those of Professor Grashof, the 
reasoning followed being substantially that given by him in his 
"Festigrkeitslehre." . .... . .-. ... . 



, ROUND PLATES. 

Let the curved line CA be a meridian curve of the middle 
layer of the plate after it is 
bent. Take the origin at O ; 
let axis OZ be vertical, and axis 
OX horizontal, and let the axis 
at right angles to ZOX be O®, 
so that z> x, and .<£ are the co- 
ordinates of any point in the 
middle layer of the plate. 

Let y denote the (vertical) 
distance of any horizontal layer 
from the middle layer of the 
plate. 

Let R = radius of curvature 
of meridian line at any point 
(x, z, <f>). 

Let R t = radius of curvature of section of middle layer 
normal to meridian line. 




Fig. 303 



Then we should have, from the differential calculus, 



R 



dx 2 



hW 



d 2 z . 
nearly, 



R T = .-± S. 



1 dz_ 

x dx 



Hence, reasoning in the same way as in the common theory of 
beams, we should have, for the strains of the layer whose dis- 



700 APPLIED MECHANICS. 

tance from neutral layer is y at point (x, <f>), provided there is 
no stress in the plane of the neutral layer, 

** - *%> h = ± x 

When there is such a stress, let the strains due to that 

stress be e Xo and €^ . 

Then we shall have 

d 2 z , v 

Hence, substituting in (i) and (2) of § 283, we have 

mE r . / d 2 z . 1 dz\l . , \ 

°* = j^ttL*S + «*, - ^ + - s)J. (3) 

mE r 1 /^ 2 2 , tn dz\~\ , \ 

°* = s^tLs + " £ *° - ^a? + *s)i (4) 

Now let us suppose the plate to be subjected, before load- 
ing, to a uniform pull in its own plane, and normal to its cir- 
cumference ; and let the intensity of this pull be/,. Then 

<r* = o-£ o = /i ; 

and hence, from (1) and (2), we have 

**o = **o = —^ , (5) 

Therefore, substituting in (3) and (4), and reducing, 

mEy ( d 2 z 1 dz\ 

wisy ld 2 z m dz\ 




STRENGTH OF FLAT PLATES. 701 

These equations express the stresses in terms of the co- 
ordinates of the points. 

Now impose the conditions of equilibrium upon the forces 
acting on any half-ring of thickness dx = d<f>. 
These forces are — 

1 °. Force exerted upon it by the outer part (f _^_x 
of the plate, 

\ 2X(T X + 2d(x<T x ) \dz. FigT7o 4 . 

2 . Force exerted by the inner part of the plate, 

— 2X(T x dz. 

3 . Force exerted upon it by the other half-ring, 
— 2<r$dxdz. 

4 . Force exerted by resistance to shear on top and bottom, 
\ (t -f dr) — r\2xdx. 

Hence, equating to zero the algebraic sum of these, and 
reducing, we obtain 

dr dr <t$ I d(xcr x ) 



dz dy x x dx 



(8) 



Now substitute for <r x and a-^ their values, and reduce, and 
we have 

dr _ m 2 Ey fd^z 1 d 2 z 1 dz\ . . 

dy m 2 — 1 \dx l x dx 2 x 2 dx) 

Integrate with regard to 7, and we have, since the quantity 
in brackets is not a function of y, 

m 2 Ey 2 /d*z 1 d*z , jt_ dz\ , 

2{ni 2 — i)\dx 3 x dx 2 x 2 dx) 



702 



APPLIED MECHANICS. 



But, whenjj/:= - (h being the thickness of the plate), t == o, 
since there is no shearing-force at top or bottom ; 
m 2 Eh 2 /d 3 z . i d 2 z i dz\ 






_(d*z i d 2 z 
8(m 2 — i)\dx 3 x dx 2 



x 2 dx A 



m 2 E(/i 2 — Ay 2 ) fd*z i d 2 z I Vs\ 
8(#z 2 — i) w# 3 x dx 2 x 2 dx) 



(10) 



This gives us; the intensity of the shearing-stress at any 
point (x, z) at distance y from middle layer ; and this is the 
intensity of the shear at that point between two horizontal 
layers, and hence also along a vertical plane through the point 
(x, z). 

Now let us take the case of a centre load P combined with 
a distributed load / per unit of area. Then shearing-force at 
distance x from centre = ■ - 

TTX 2 p 4" P, 

this tending to shear- out a circular piece of radius x. Hence 
we must have this balanced by the whole shearing resistance 
on the surface subjected to shear; 



it 
2^xj%dy = 



TTX 2 p + P 



h 



Now substitute the value of r from equation (9), integrate, 
and reduce, and we obtain 



dx 3 x dx 2 x 2 dx m 2 E]fi 



~{ px+ S)- (i2) 



STRENGTH OF FLAT PLATES. 7O3 



Hence, for the intensity of the shearing-force, we have 

4 A 3 \ irXj 



r 

4 



This gives the intensity of the shearing-force at any point of 
the plate. 

Next, to find its deflection, or the equation of the meridian 
line, we have, from (12), 



d ld 2 z 1 dz\ _ 6(m 
dx\dx 2 x dx) m 



M^S) 



d 2 z 1 dz 6(m 2 — 1) x 2 6 Cm 2 — 1) P. 

1 = -p - —\og a x + c 

dx 2 x dx m 2 Eh?> 2 nfEfa ir 

d 2 z , dz 6(m 2 - 1) x* 6(m 2 — 1) P , 

X | = -p— — X log, x -h ex, 

dx 2 dx m 2 Ete 2 m 2 Eh* ir 



But 



d 2 z dz_ __ d_l dz\ _ 

dx 2 dx dx\ dx) i - — . 

hence, integrating, we have 

dz '_ 6{m 2 — 1) x 4 

X ~dx~~~ m 2 Eh> ^8 

6(m 2 — 1) Px 2 , , 6(m 2 — 1) x 2 , ex 2 ■ , -. . 

- •-.:,, ■ - -log,* ■+ ,--, ' —'+—■+ d. (14) 
m 2 Efc 7T 2 m 2 Eh* 4 2 

Hence, dividing through by *", and integrating, 
6(m 2 — 1) * 4 

2= — p— . : ; - — _' 

m*Eh* 4 

6(;« 2 — 1) Px 2 ,, ■. £# 2 , .. , . 

2i?lx ( lo &* - 0-+ — + ^log,* + e; (15) 

m^LLn* ir 4 4 

and this is the meridian line of the surface, the constants c, d, 
and *? being as yet undetermined. 



704 APPLIED MECHANICS. 

This is as far as we can proceed before taking up special 
cases. 

(a) Full Plate, — When the plate is full, the slope becomes 
zero, for^r = o; therefore (14) gives- us 

d — o, 

and in this case (15) becomes 

_ 6(m 2 — 1) x* 
m 2 Efa 4 

6(m 2 — 1) Px* ,. ■ , ex 2 , , ' 
agz, ( lo & x - + — + * (16) 

dz d 2 z 

And, substituting for z, — and — — ,*their values in (1) and (2), we 
ax ax 2 

obtain 

*<* = z^rr* + \l ~^Ehr p - nh (I7) 

^ = ^_ZJ + (I 6 ^-^ P - C X; ( l8 ) 

9 W ° \8 W 2 ^ r 2 f V ^ 

and (13) gives 

T = J ~* -^ (I9) 

(a) Uniformly Loaded, no Centre Load.—P =. o ; 

6(m 2 — 1) x* ex 2 , , v 

m 2 Eh* 44 

But when .ar == r, £ = o ; 

_ 6(w 2 — 1) r* cr 2 
m 2 Eh* 32 4 

/6(^ 2 — 1) x 2 -f ?* 2 \/> 2 — oc 2 \ ,«•;/* 

••• s = (^-^-^-8- - 'A— r~/ (2I) 



STRENGTH OF FLAT PLATES. 705 

(/?) Supported all around. — When x = r, a- x = <t Xq == / r for 
all values of ^.* therefore, from (6), 



w 



\dx 2 / r r\dx/ r 



dz d 1 ''2 

and, substituting the values of — and — — as determined by 

ax ax^ 

differentiating (21), we have, after reducing, 

c _ 3 ( m ~ 0(3"* + 1 ) P r * 
2 m* Eh? 

Hence equation of meridian line is 

Efr\m + 1 ) v 1 \ / 



z = ^ m 



1 6 #/ 2 



Hence we have maximum deflection by making x = o ; 

. ,J3 O ~ i.) (5* + 1) /** . v 

And, substituting in (17) and (18), we obtain, after reduction, 






-a + - — — £ r T r - * x }^ (24) 

4 m 2 h> ( #z -+- 1 \ 

-£«♦ = — a + - — — £ r r r * - ^ f* ( 2 5) 

m 4 m 2 fa ( m + 1 j 

But, in a plate supported all around, p t = o ; and then the 

maximum value of either one occurs when y = -, and hence 

2 

^ = 3 (^-i)(3^ + i)^ (26) 

8 #/ 2 # v ' 



7-06 APPLIED MECHANICS. 

On the other, hand, t becomes greatest when x = r and 
y = o. Hence 

Max^r = ^/; 

and, if €l represent the maximum strain due to this shearing- 
force, we have 

Max (£,,) =Y^+jVmax£r) = * *L±-i »£ (2?) 
\ m J 4 m h 



RESULTING FORMULA FOR PLATE SUPPORTED ALL ROUND. 

„ 2 (m — \)(%m -f i) r 2 \m-\-\r- 

Max E* Q = | ^ n6 > -p or * -/, 

8 m 2 h 2 4 m h 

whichever is greatest. 

3 {m — i)(5^ -+• i) pr* 



z n = 



1 6 m 7 - Eh* 



PLATE FIXED AT ENDS. 

Equation (20) applies to this case" also. 

Now, when x == r> — ■= o ; 
ax 

.-. c = 2 =t-/^ 

2 m 2 Eh> 



-- 16 .*» £/^ V , > \ / 

Hence greatest deflection S 

_ 3 pi 2 ~ i pr 4 



THICKNESS OF PLATES. 707 



and 



^f-.^tt^fe^ (29) 

;^i^fi^ifl:^ (3o) ... 

When p l is positive or zero, then E* x is maximum for x = o 
7 = -, and for ^ = r, y = ; and iTe,* is maximum for 

2 2 

j = oj = -: and the maximum value of E^ is equal to first 
maximum of Ee x . We have 

First max^ = — p x + | -^ _/, (31) 

m 8 w 2 >£ 2 

o j 77 #z — i , 3 ;« 2 — i r 2 . x 

Second max JSe x = A -f p. (7.2) 

m r 4 m 2 h 2 * vo y 

Hence the second is the real maximum. 



RESULTING FORMULA FOR PLATES FIXED AT THE ENDS. 

_ , _ m — i , -? m 2 — 1 r 2 . 

MaxE^ = $ + 2 — -p } 

m 4 m 2 h 2 

3 w 2 — 1 /r 4 
*° "" 16 m 2 ~Eh\ 



For /j = o, 



,. „ * m 2 — 1 r 2 M 

Max^o = 2 — p. 

4 ;« 2 /* 2 



§ 299. Thickness of Plates. — Grashof advises the use of 
3 as value of m. If this be adopted, we should have, for the 
proper thickness of round plates, 

Supported. Fixed. 

-=V| wg. 



708 



APPLIED MECHANICS. 



where h = thickness, r — radius, / = pressure per square inch, 
and / == working-strength per square inch. If, now, we use a 
factor of safety 8, and use as tensile strength of cast-iron 20000, 
of wrought-iron, 48000, and of steel 80000, we should have : — 





Supported. 


Fixed. 


Cast-iron . . . 
Wrought-iron 
Steel .... 


h = O.O1825 lorSp 
h — 0.0117850?-^ 
h = 0.0091 2d>iryp 


h = 0.0 1 63300/- yp 
h = 0.0105410/-^ 
h = 0.0081649?-^ 



§ 300. Rectangular Plates. — Refer the plate to rectangu- 
lar axes, as before, OZ, OX> 0& ; the origin being at the middle 
of its middle layer. 

Let y = distance of any point in the plate from the middle 
layer. 

Let p x be the radius of curvature of a normal section par- 
allel to OX at the point (x y z, <f>). 

Let pj, be the radius of curvature of a normal section par- 
allel to 0® at the point (x, z, </>). 

Then we shall have, by the principles of the common theory 
of beams, 



€ x ± 



Px 



H = H Q ± 



P<t> 



where € x and e^ o are the strains of the middle layer in the 
directions OX and O® respectively. 

Moreover, from the Differential Calculus, we have 



^■> (!?*<$)" 



p * ~" d 2 z , x , d 2 z , , d 2 z • ' 

— - COS 2 \ -h 2- COS A. COS U -f- -— COS 2 /A 

dx 2 dxd<f> d<f> 2 



RECTANGULAR PLATES. 709 

where A. =s angle between normal and z axis, and ^ = angle 

dz dz 

between normal and x axis. But — - and — being the slopes, 

dx d<$> v 

and hence small, we shall have nearly 

cos A = 1, cos /a = o, 



1 d 2 z 1 d 2 z 
9x doc 1 ' p$ dcfi 2 




d 2 z 


(1) 


d 2 z 
9 ^° d<& 


(2) 



Hence (1) and (2) of § 283 give us 

mE , . mE i d 2 z d 2 z) 

mE ' N mE ( d 2 z , d 2 z ) 

vi 2 — 1 ° ° m 2 — 1 (dx 2 d<f> 2 ) 

And, if or^, cr^ o , denote the stresses in the middle layer, we 
shall have, since 



U ,,,,2 






m 

m 2 E [d 2 z . 1 </ 2 2 



OV = CTj-. — 



\d 2 z I tf 2 2 ) 
/w 2 - /{dx 2 md$ 2 )' ^ 

m 2 E [1 d 2 z , d 2 z) , . 

Now, if I and >? denote the increments in # and <f> respec- 
tively due to the load, we shall have 



£ = I *xdx = X€ Xq - y 



dz d£ _ ^ 2 2 

</# */<£ dxd<fi' 

, dz drj d 2 z 



</<£ dk dxd<f> 



710 APPLIED MECHANICS. 

But 



hence 



dxd§ 



Equations (3), (4), and (5) are the expressions giving the 
stresses on two planes at right angles to each other, parallel to 
OX and 0$ respectively. Hence we have a case of stress on 
two planes at right angles to each other, and we are to find the 
principal stresses : we thus have — 

i°. Normal stress on x plane, <r x . 

2 . Shearing-stress on x plane, t^. 

3 . Normal stress on <£ plane, o-^,. 

4 . Shearing-stress on <f> plane, t^. 

Hence, if we denote by <r I and o- 2 the maximum and mini- 
mum principal stress, we have (§ 290) 

<r* = K°* + o**) 4- i)/(<rx + °>) 2 + A^xf, (6) 



<r 2 = J(o> + <ty) — $\!(<r x + <r*) 3 4- 4 W* i (7) 

and hence, if e x and e 2 denote the strains in the directions of 
the principal stresses, 



o- 2 m — 1, 
m 2m 



m 4- 1 



+ -^-V(cr^ + <r*)» + 4 r** a , (8) 



_ <r x m — \ f N 

^€ 2 = (T 2 = (tf^ + Oty) 



W 4- I 



; V(o* 4- <r$Y + 4^ 2 ; (9) 



RECTANGULAR PLATES. /I I 

and for the strain 4, parallel to 0Z> we have 

<r x -f 0-4, 

In order to use (8), (9), and (10), however, we must know 
°V, °>, and t x4> ; and for this purpose we must know the equa- 
tion of the middle layer after bending. For this purpose, apply 
the equations (1), (2), (3), of § 281 to any particle dxdfydz in 
the interior of the body. We have then, X = Y = Z = o. 
Therefore 



dcr x dr xz dr x $ 

dx ~*~ dy + d<j> ~~ ° 


dr xz lda x dr x A 


dcr^ dr X( f, dr$ z 

~dj> + ~~dx + ~dy '' 


dr^ z fda-fr dT x< A 
dy :Z \dcj> "*" ~^/' 



^"g , dr<i> z d- X2 _ 
dy d<j> dx , . 

Therefore, making use of (3), (4), and (5) with the above 
conditions, we deduce 

dcr x m 2 Ey (d 3 z 1 d 3 z \ 

dx m 2 — \dx* m dxd<f> 2 )' 

d<T<$> ' m 2 Ev IdH 1 d*z V 

~^~ ~~ ;^ _ iV^£3 mdx 2 ^)' ^ 12) 

dr xy _ viEy d*z . 

~Tx ~ ~~ m + 1 ^ 2 ^>' 

dV^ _ jnEy__ _dH_ ( 

d<f> m -f- 1 dxd<^ 

dT xz __ ??i 2 Ev Id^z d*z \ 



tfV<^ _ m 2 Ev /dz* d*z N 

*/y w 2 — i\^v 3 dx*dyj 



(16) 



712 



APPLIED MECHANICS. 



Hence, by integrating (15) and (16), we have 
m 2 Ev 2 (fcz , d*z \ 



?XZ = 



T<}>z = 



2 (;// 2 - i)\dx* dxd<j> 2 ) H 
m 2 Ev z IdH . d*z 



2{m 2 — i)\d<f>3 dx 



h 






But when v = -, r+ z = r xz = o ; 



and 






c 2 = — 



\ 2 Eh 2 IdH 



('- 



dH 
8(m 2 — i)\dx* ' dxdcf* 

m*Eh 2 IdH 
8(m 2 - i)\d$* ' dx 



dH \ 

7 x 2 d<f>) 



and 



Hence 



m 2 E I dH dH\ !tf_ __ fr\ 
r * z ~ m * - \dx 2 d<}> dp) \ 2 "" 8 / 



m 2 E I dH dH \l^ _ &\ 
~ m 2 — 1W 3 dxd<t>*J\2 " 8 / 



dr^ z _ 
d<f> tn 

dr x 
dx 



m 2 E 



I dH dH\/v 2 _ k 2 ^ 

~i\dx 2 d<f> 2 dp)\2 ' 8 A 



m 2 E IdH 



m 2 — i\dx* dx 2 d<f> 2 

Now we have <r z = p y where / is the intensity of the load ; 
therefore the third equation gives us, on integrating between 

the limits - and , 

2 2 



2 2 

h h 

J_*d<t> JJL dx J 



.-. / + 



RECTANGULAR PLATES. 713 



m 2 - i(\6 ~&)\dx* 2 dx*d<p ~d&)_h\ " 



m*E ( d*z d*z d*z\f fa __ fa\ _ 

. ^ . ^ 4 s , d*z __ i2(m 2 - i)p . (i) 

dx* dx 2 d<j> 2 d$* m 2 Efa 3 K 1) 



and this is the differential equation of the surface, and should 
be integrated in each special case. 



INDEFINITE PLATES WHICH ARE FIRMLY HELD AT A SYSTEM OF 
POINTS DIVIDING THEM INTO RECTANGULAR PANELS. 

Let the sides of the panels be 2a and 2b. Assume the 
origin at the middle of the panel, the axis of x being parallel 
to 2a, and the axis of y parallel to 2b. We shall in this case 
have the following conditions ; viz., — 

dz 

(a) — - = o f or x ' = ±a and all values of <f>. 
ax 

dz 

(b) — - = o for <f> = dtb and all values of x. 

d<t> 

(c) z = o when x = zka, <f> = ±b. 

(d) If we develop the value of z in powers of x and $, there 
must enter only even powers of x and <£, since the value of z 
remains the same when we put — x for x, or — <j> for <£. 

Now, if we write 

z = A + Bx 2 + C4> 2 -f Dx 2 4> 2 + Ex* 4- F<p 

+ Gx 6 + Hx*<p + Kx 2 ^ + Lcj> 6 + Mx*, etc., 

the above conditions will be fulfilled : — 



7 14 APPLIED MECHANICS. 



i°. By making all the co-efficients after the fourth, each zero. 
2°. By making D = o, therefore writing 

z = A + Bx 2 + C<f> 2 + Ex* + F<p. 



Now 



^ = 2 Bx 4- 4 ^ 3 , ^ = 2 ^ + 4 ify3, 

## a<p 



/. 2^^ + 4^ = o, iCb -f 4^3 = o, 

and 

o = A 4- Ba 2 + C2 2 + ^ 4 4- ^ 

.\ ^ = -2Ea 2 , C = -2Fb 2 , 

:. A = 2^4 4- 2Ffi* - Ea* - Fb* •= *&** 4 F6\ 

Hence the equation becomes 

* == Ea* + FM — 2^a 2 ^ 2 — 2/tf 2 <£ 2 + -£*♦ + 7^4 

= E(a 2 — x 2 ) 2 4- <F(£ 2 ~ </> 2 ) : 

— = — $Ex{a 2 — x 2 ) = 4^^ 3 — ^Ea 2 x 
dx 



also 



— = 12EX 2 — 4Ea 2 } 
dx 2 



— = -4F$(l> 2 - <f> 2 ) = 4F<I>3 - 4Fb 2 $ 
d<fi 



.'. — = i2F<j> 2 - 4i^ 2 , 
// 3 s; f ^ 4 s ^ ^4 2 



dx 2 d$ dx 2 d(j> 2 dxdcf> 2 dx 2 d<j> 2 

d*z „ dH „. </ 4 s ^ 

— = 2 4 ^, _ = 2 4 ^, - =2 4 ^, _ = 2 4 ^ 

., 2A{E + ■*) = i2(m2 - i) ^ .•: . .* t'^= (; * 2 ~ i}/ . 

Hence equation of the middle layer is 

% = E(a 2 - x 2 ) 2 4- TW - of) 2 ) 2 , where E + F = ("* 2 ~ ')/ (l8) 

2m 2 Eh* 



RECTANGULAR PLATES. 



715 



Now, in the case of an ordinary beam fixed at both ends, 
and loaded uniformly with / lbs. per unit of area, if b is the 
breadth, we have : — 

i°. The points of inflection are at a distance from the middle 

equal to -=, where a is the half-span ; and 
V3 
2°. The bending-moment at a section at a distance x from 

the middle is^— ( x 2 \ when x <-=, andv-f^ 2 — — ) when 

.2\3 J V3 2\ 3/ 

x > — ; therefore the value of z is found from the formula 

or 

JSfcJo Jo \3 / 

Either one, when integrated, gives for z the value 

Hence in the flat plate, if b = o, the values of E and i 7 must 

-(tf 2 — x 2 ) 2 



be such that the formula shall reduce to # = - _ 

2/sY* 3 

when b = o. Now, it does reduce to # = 2T(« a — x) 2 . There- 
fore E must be such a function of a and £, that, when b = o, it 

So likewise F must be such a function 



shall reduce to 



2Eh* 



of # and £, that, when a = o, it shall reduce to 



2^ 3 



Suppose, 



then, we put 



2 Eh} 



and 



2Efa 



since these functions fulfil the above conditions. 



J\6 APPLIED MECHANICS. 



Now we have 



E + F = ( m2 -^ 
2tn 2 Efr 



2Efc V ' 2W^3 









2tn 2 Ete 

(m 2 + i)/ 
2tn 2 Efo{a n + £*) 



_ 1 ■ wafts 

m 9 ° J dx 2 



T? 1 T?d Z 

EtA = o-a ov — yE——. 

9 9 ° m ° J d<f> 2 

d 2 z d 2 z 

Hence, substituting for — — and -r- their values, and observ- 

dx 2 d<f> 2 

ing that 

' e x is greatest for x = ±a, y = ±-, 

2 



ty is greatest for <£ = ±£, j/ = ±-, 

2 

we obtain 

I #/ 2 2 

max (E* x ) = <r, - -o> ± 2 -/, (19) 



£» —a n 

1 w 2 ^ 2 

max (E H ) = <r* - -a Xi ± 2 an + in jj. (20) 



RECTANGULAR PLATES. ?I? 



These may be written as follows : 



m\a) 



i 
77 i , m 2 \aj a 2 ^ 
max Ee x = a Xo - -cr* o ± 2 ^— . -/, (21) 

w 

l/^V 

z7 1 m\b) b* '• 

maxAcA = ov. <r~ ± 2 \ - —/. (22) 

We have also, by substituting for E and F their values in 
equation (18), 

, a n —b 

p \ m 2 



b n —a n 

- * 2 ) 2 + . ,1 & - VY ■ 



(a 2 - x 2 ) 2 H — (b 2 - <£ 2 ) 2 f . (23) 

In these results the exponent n is undetermined, and we 
have no means of determining it in the general case. We only 
know, that, since the deflection must increase for a decrease in 
x and <f>, therefore we must have, whenever a > b, 

a\ n , 2 lose m 

i) <m ••■• n< rP\ 

This leaves the general case indeterminate ; but a common 
practical case is not subject to this indetermination, i.e., the 
case when a = b, for then 



©"-©"-- 



whatever the value of n; and hence equations (21), (22), and 

(23) give 

/ t? \ 1 , m 2 — 1 a 2 M , N 

max (E* x ) = <r, o - -^ o ± — — -A (24) 

max E H = <r« - -<r Jo ± _j_ A (25 ) 



JlB APPLIED MECHANICS. 

z = "^1 j£_j f J _ Pp + (# _ ^jif, ( 26 ) 

m 2 AfEte 

and 

w 2 — i a* < . v 

maxz s= ^tt-A (27) 

w 2 2^^^ v " 



FORMULA FOR THE SHEETS OF A LOCOMOTIVE FIRE-BOX. 

In this case we have a = b; hence (24), (25), and (27) 
apply : and if we write, with Grashof, m == 3, they become 

max (Ee x ) =cr x -- o-^ + - ^J, (28) 

3 9" 

max (E H ) = o> - - <r x + - -/, (39) 

3 9 " 

max(s) = ^-|^. (30) 

9 z?^ 3 

Now, in the case of the horizontal sheets;, o-^ = cr^ = o, 

and we have 

, „\\ 8 a 2 . /' x 

max (^^) = v -A (31) 

maxW = ^- ^ ^ 

In the case of the vertical walls, inasmuch as these have to 
resist the steam-pressure in a vertical direction, the inner one 
is called upon to bear compression, and the outer tension, in a 
vertical direction. If /is the length of the outside of the fire- 
box, and /, its breadth, we shall have for the outer plate, taking 
axis of x vertical, 



RECTANGULAR PLATES. 71$ 



and for the inner plate, if / and // are corresponding dimen- 
sions of inside of fire-box, 



ir t p 

<TA = O. 



2(/ / + ll)ti 

And, by making these substitutions in (28), (29), and (30), we 
obtain our formulae. 



RECTANGULAR PLATE FIXED AT THE EDGES. 

For this case Grashof deduces the equation of the middle 
layer as follows : — 

i°. This equation must be a function of x and </>. 

2°. If 2a and 2b are the sides of the plate, this function 
must become 

(a) When b = oo for all values of <f>, 

z = -4rr( a2 — * 2 ) 2 - 
(/?) When a = oo for all values of x, 

*Ehy v ' ' 

because the plate then becomes a beam fixed at the ends. 
The function that will satisfy these two conditions is 

Zrr , P (* 2 - * 2 ) 2 0* 2 - ¥Y (l) 

2Eh* a* + b*> • ... V 



From this he deduces for max z, when # ■■ =t </» = o : 



max 2 = - -£~ -.- (2) 



720 APPLIED MECHANICS. 

From (i) he deduces 

d 2 z _ _ 2p (a 2 - zx 2 ){b 2 - <t> 2 ) 2 ( ., 

dx 2 Eh* a< + M ' U; 

d 2 Z _ 2p (g» - *»)»(* - 3 <ft 2 ) 

dty 2 " Eh* a* + t* ' W 

d 2 z _ 8/ (a 2 - * 2 )*(^ - <j> 2 )<f> 
dxdcf> ~ ^^ a * + £ 4 » ^ 5) 

d 2 z Ap a 2 b* f , , ,, x 

max — = -*£. -, for x = ±a, = o, (6) 

max— = -|£- , for <£ = ±J, x = o, (7) 

dty 2 i?>& 3 a 4 -f- <? 4 



</ 2 s 
max 



= 3g-A ^ for * 2 = V, ^ = U 2 , (8) 



dfcdty 27 ^ tf4 + £4 3 3 

these corresponding to the points of inflection of a loaded 
beam fixed at the ends. 

Hence (1), (2), and (5) of § 300 give 

maxzj.e x = ov cta ± —A (9) 

I 2# 4 ^ 2 

max {E H ) = <r, o - -^ o ± -j-j-^j -A (to) 

, x 8 m 2a 2 b 2 ab M , x 

max (t 2 ) = — /. (11) 

At the places where * x and c^ are greatest, 

T Z = O. 

At the place where t s is greatest, 

<?x *s ov o , os, = cr„ o . 

Hence it is either (9) or (10) that gives us the suitable 
formula to use in any special case. 



EXAMPLES OF THEORY OF ELASTICITY. 72 1 



EXAMPLES OF THEORY OF ELASTICITY. 

i. It has been sometimes proposed to use oblique seams in a boiler- 
shell. Assume the seams at an angle of 45 with the axis of the boiler, 
a pressure of 100 lbs. per square inch of the steam, and a diameter of 
4 feet. Find the tension per inch of length of seam, and its direction. 

2. Given a shaft carrying 80 HP, and running at 250 revolutions 
per minute. Suppose the driving-pulley to be at the middle of the 
length, this being 6 feet, and given that the ratio of the tension on the 
tight side of the belt to that on the loose side is 3.75. Find the proper 
size of shaft, assuming 10000 lbs. per square inch as the working-strength 
of the iron. 

3. What should be the thickness of a flat plate to bear 150 lbs. 
pressure per square inch, and stayed at points forming squares 8 inches 
on a side, the plate being of wrought-iron, working-strength 10000 lbs. 
per square inch. 

4. Find inner radius of a hydraulic press to bear 1500 lbs. per 
square inch, given outer radius =18 inches; material, cast-iron; ten- 
sile strength 20000 lbs. per square inch. 



INDEX. 



PAGE 

Acceleration 75 

Angular momentum 106 

Arches 602 

conditions of stability 612 

correcting joints 623 

criterion of safety 613 

criterion of stability 628 

elastic 637 

general remarks 635 

linear - 601 

line of resistance 607 

line of resistance determined by two 

points 616 

modes of giving way 602 

Scheffler's method 617 

true line of resistance 615 

unsymmetrical arrangement . . . 629 

Atwood's machine 79 

Axes of symmetry of plane figures . .115 

Bar-iron, tests by Kirkaldy 357 

Barlow, modulus of elasticity of wrought- 
iron 355 

Bars and shapes of wrought-iron, brands, 381 

Bars, steel, tests by Kirkaldy . . . .411 

Beams, assumptions of common theory . 258 

cross-section of uniform strength . 288 

deflection of 289 

deflection with uniform bending-mo- 

ment 296 

fixed at the ends 302 

load not at middle 297 

longitudinal shearing 309 

mode of ascertaining dimensions . . 282 
mode of ascertaining stresses . . .281 



PAGE 

Beams, modulus of rupture 283 

moments of inertia of sections . . 265 

oak 536 

position of neutral axis 259 

principles of common theory . . .257 
rectangular, slope and deflection . .301 

resilience of 296 

shearing-force and bending-moment, 262 

slope and deflection 291 

slope and deflection, special cases . 292 
slope and deflection under working- 
load 300 

spruce 527 

table of deflections and slopes . . 295 
table of shearing-forces and bending- 

moments 264 

timber, strength and deflection . .523 

timber, time tests 534 

uniform strength 286 

variation of bending-moment with 

shearing-force 307 

white-pine 537 

working-strength 283 

wrought-iron, strength and deflec- 
tion 397 

yellow-pine, strength and elasticity . 532 

Beardslee, effect of rest 370 

reduction in rolls, wrought-iron . . 369 

shape of specimen 368 

tensile limit 368 

tests of wrought-iron 367 

Bending and torsion combined .... 690 

and twisting 328 

Bending-moment 175 

in beams 262 

723 



724 



INDEX. 



PAGE 

Bending-moments, graphical representa- 
tion 277 

Boiler-plate, brands 380 

Boiler-plate tests by Richards .... 378 

Bollman's truss 209 

Bow's notation 143 

Brands of wrought-iron 380 

Breaking-strength • . . .235 

Bridge columns, tests for Clark, Reeves, 

& Co 383 

Watertown-arsenal tests 389 

Bridge-pins 443 

Bridge-trusses 174 

actual shearing-force 193 

compound 198 

concentration of loads at joints . .193 

counterbraces 190 

diagonals . 190 

examples 176,284 

general formulae 209 

' general remarks 209 

method of sections 1 74 

1 shearing-force and bending-mcment . 175 
steps in determining stresses under 

fixed load 176 

vertical posts . 192 

with vertical and diagonal bracing . 188 

Building-stones . . ■ 550 

Buttress,, stability 605 

Cast-iron, columns 344 

composition and characteristics . . 332 

effect of skin 351 

extension and compression, Hodgkin- 

son 339 

list of experimenters ...... 336 

modulus of elasticity 340 

modulus of elasticity, Rosset . . . 344 
tensile and compressive strength . . 336 

tests by Rosset 343 

tests by Wade 341 

transverse strength 349 

Catenary 596 

transformed 599 

Cement mortar 553 

Centre of gravity 211 



PAGE 

Centre of gravity, examples 216 

of a line 215 

of a slender rod 214 

of flat plate 213 

of homogeneous bodies 213 

of plane area 214 

of solid bodies 226 

of symmetrical bodies 227 

of system 211 

Pappus's theorems 224 

Centre of percussion 120 

of stress 253 

Centres of percussion and of oscillation, 

inter changeability 123 

Chain cable 482 

Chain or cord, loaded 591 

Cold-rolling 408 

Collision 123 

Columns, cast-iron 344 

Euler's rules 318 

Gordon's formulas 3115 

Hodgkinson's rules 321 

oak 50^ 

strength of 314 

timber 499 

white-pine 511 

wrought-iron 383 

yellow-pine 503, 516 

Components of velocities of, and forces 

acting on, a body 82 

Compound bridge-trusses 198 

Compression, direct 243 

of timber . 499 

Compressive strength of cast-iron, Hodg- 

kinson 336 

of wrought-iron 382 

Wohler's experiments 244 

Continuous girders 555 

distributed and concentrated loads . 582 

examples for practice 590 

loads concentrated 575 

loads distributed 556 

Contraction of area, Kirkaldy .... 362 

Cord or chain, loaded 591 

Cord with load uniformly distributed 

horizontally 594 



INDEX. 



725 



PAGE 

Counterbraces 190 

Couple, composition of, in inclined planes, 59 

Couples, composition of 57 

effect on rigid body 55 

effect when forces are inclined to 

rod 54 

measure of rotatory effect . . . . 53 

moment of 53 

effect on rigid rod 51 

representation by a line 59 

resultant with single force .... 61 

Crystalline fracture, Kirkaldy .... 362 

Cylinders, thick hollow, strength of . . 694 

thin hollow 241 

Deflection of beams .... 289, 291 

Domes 649 

Dynamics 75 

Eccentric load on timber columns . . 506 
Elasticity, immediate modulus for timber 

beams 536 

modulus, cast-iron 340 

modulus for use with timber beams . 535 

modulus of 230 

modulus, Rosset 344 

modulus, wrought-iron, Hodgkinson, 353 
modulus, wrought-iron, Barlow . .355 
special modulus, wrought-iron, Ros- 
set 377 

tensile modulus, wrought-iron, Water- 
town Arsenal 374 

theory of 658 

Ely, rules for strength of timber posts . 521 

Energy 78 

Equilibrium curves 591 

Euler's rules for columns 318 

Eye-bars, steel, Hill 415 

steel, Watertown Arsenal . . . .417 
wrought-iron 376 

Factor of safety for iron and steel . . 425 

timber . 522 

Fibrous fracture, Kirkaldy 362 

Fink's truss 207 

Floors, timber 546 



PAGE 

Force 3 

and momentum, relation between . 11 
applied at centre of gravity of rigid 

rod 51 

applied to rigid rod, not at centre . 46 

centrifugal 8[ 

centrifugal, of solid body .... 85 

characteristics of 16 

criticism of definition 6 

definition of 8 

deviating 82 

external 9 

intensity of, distributed 40 

measure of 5, 9, 76 

moment of 30 

relativity of 9 

resultant of, distributed 40 

single, at centre of rigid rod ... 43 

Forces, centre of system of parallel . . 38 

composition of 21, 23 

composition of parallel . . . . 37, 62 

composition of parallel, in a plane . 36 

composition of two 30 

co-ordinates of centre of parallel . . 39 

decomposition of 19 

distributed 39 

effect of pair on rigid rod .... 50 

equilibrium of 28 

equilibrium of, in a plane .... 69 

equilibrium of, in space . .... 73 

equilibrium of parallel 37 

equilibrium of three parallel ... 31 

moment, causing rotation .... 49 

normal and tangential components . 80 

parallelogram of 16 

polygon of 23 

resultant of any number of parallel . 35 

resultant of, in a plane 66 

resultant of, in space 70 

resultant of two parallel 33 

statical measure of 12 

triangle of . . . 19 

Foundry iron 333 

Frame, triangular 141 

isosceles triangular 143 

polygonal 145 



726 



INDEX. 



PAGE 

Frames of two bars 138 

Frames, stability 140 

Framing-joints 546 

Friction of blocks . .603 

Funicular polygon 147 

G, relation to E 674 

G, value of 673 

Girder, greatest stresses 183 

Girders, continuous 555 

distributed and concentrated loads . 582 
examples for practice ...... 590 

loads concentrated 575 

Gordon's formulae 315 

Gordon's formula, applicability to bridge 

columns 383 

modified forms 385 

Grade, effect on tractive force . . . .100 
Gravity, centre of 42,211 

Half-lattice girder : travelling-load, 

greatest 182 

Hammer-beam truss 166 

wind pressure 169 

Harmonic motion 102 

Hill, steel eye-bars 415 

steel plates 423 

Hodgkinson, tensile strength and elas- 
ticity of wrought-iron 353 

Hodgkinson's rules for columns . . . .321 
tests of cast-iron columns .... 344 
tests on tension and compression of 

cast-iron 337 

Hooks, strength of 312 

Impact ............ 123 

central 124 

co-efficient of restitution 125 

co-efficient of restitution by experi- 
ment 132 

elastic 127 

imperfectly elastic 1 29 

inelastic 126 

oblique 134 

of revolving bodies 136 

special cases of elastic 128 



PAGE 

Impact, special cases of imperfect elas- 
ticity 132 

special cases of inelastic 128 

velocity at greatest compression . .125 

Kirkaldy, riveted joints 477 

tests of bar iron 357 

tests of iron plate 360 

tests of rivet iron 359 

tests of steel 408 

tests of wrought-iron 356 

Kirkaldy's conclusions 361 

Launhardt's formula 238, 429 

Load, sudden application of 236 

Longitudinal shearing of beams . . . 309 

Mass, measure of 10 

unit of 13 

Materials, strength of . 230 

Metals and alloys other than iron and 

steel 486 

Modulus of elasticity 230 

approximate values 235 

Modulus of rupture, tor beams .... 283 

Moment of deviation 108 

Moment of inertia 106 

Moments of inertia about different axes, 113 

components of .117 

equal values of 116 

examples 118 

of Phcenix columns 276 

of plane figures about parallel axes . no 

of plane surface 107 

of sections 265 

of solids around parallel axes . . .117 

polar, of plane figures in 

principal 114 

Momentum n 

Mortar (cement) 553 

Motion and rest 2 

Motion, Newton's first law of . . . . 4, 9 

Newton's second law of 13 

on curved line 95 

on inclined plane 92 

relativity of 1 



INDEX. 



727 



PAGE 

Motion, under influence of gravity . . . S7 

uniform 76 

uniformly varying 76 

uniformly varying rectilinear ... 87 

Motions, parallelogram of 15 

polygon of 15 

m, value of 673 

Nails in one pound 151 

Neutral axis of beams 259 

Notation, Bow's 143 

Oak columns 504 

Pappus's theorems 224 

Pendulum, cycloidal 99 

simple circular 97 

Pig-iron 332 

Plate-iron, tests by Kirkaldy 360 

Plates, flat, strength of 699 

steel, Hill 423 

steel, tests by Kirkaldy 411 

thickness of 707 

Polygonal frame 145 

Projectile, unresisted 89 

Punching and drilling plates 453 

effect on steel, Hill 424 

Radius of gyration 107 

Rankine, strength of timber 491 

Reduction in rolls, Beardslee, wrought- 

iron 369 

Resilience of a beam 296 

of tie-bar 237 

Resistance, line of ........ 607 

line of maximum and minimum . .611 

to direct compression 243 

to shearing 246 

true line of 615,647 

Rest, effect, Beardslee 370 

Richards, tests of boiler-plate .... 378 
Rigid bodies, rectilinear transferrence of 

force 29 

rotation of 105 

statics of 29 

Riveted joints 445 



PAGE 

Riveted joints, Kirkaldy 477 

proportions given by Box .... 447 
proportions given by Unwin . . . 449 
proportions given by Wilson . . . 447 

punching and drilling 453 

Watertown-arsenal tests 455 

Rivet-iron, tests by Kirkaldy 359 

Rodman, strength of timber 495 

Roofs, estimation of load 153 

weight of materials 151 

Roof-trusses 138, 156 

determination of stresses . . 150, 155 

distribution of load 153 

examples 173 

general remarks 162 

with loads at lower joints . . . .161 

Rope, wire 484 

Rosset, special modulus of elasticity of 

wrought-iron 377 

tests of cast-iron 343 

Rupture, modulus for beams 283 

Scheffler's method with arches . . .617 

mode of correcting joints .... 623 

Scissor-beam truss 169 

without horizontal tie 170 

Seasoned columns 505 

Seasoning, effect on timber 548 

Semi-girder, greatest stresses 183 

Shafting, strength of 323 

Shafts, transverse deflection 327 

under combined torsion and bend- 
ing 690 

Shape of specimen, Beardslee .... 368 

Kirkaldy 365 

Shearing-force 175 

actual, for bridge-trusses 193 

in beams 262 

Shearing, resistance to 246 

Shearing-strength of iron and steel . .437 

Shearing of timber 547 

of timber beams 537 

Skin of cast-iron, effect on strength . . 350 

Slating, weight 151 

Slope of beams 291 

Slotted cross-head 102 




728 



INDEX. 



PAGE 

Snow, weight of 151 

Spruce beams 528 

Stability of a buttress 605 

of an arch 612 

of position 605 

Steel, composition, kinds, and character- 
istics 404 

effect of cold-rolling 408 

effect of temperature 408 

eye-bars, Hill 415 

" eye-bars, Watertown Arsenal . . .417 

list of experimenters 410 

plates, Hill 423 

tensile strength and elasticity, Water- 
town tests 412 

tests by Kirkaldy 411 

torsional strength 438 

transverse strength 425 

■ wire 4S3 

Stone 550 

Strain 230 

resultant 661 

Strains 658 

in terms of distortions 661 

relation to stresses 669 

Strength, breaking and working. . . . 235 

of columns 388 

of hooks 312 

of materials 230 

of materials, general remarks . . . 329 
of shafting 323 

Stress 230 

centre of • . . . 253 

compound 665 

converse of ellipse 688 

ellipse 685 

graphical representation 251 

intensity of 250 

relation to strains 669 

simple 664 

tangential 665 

uniform 254 

uniformly varying 255 

uniformly varying, amounting to a 
statical couple 256 

Stresses 663 



PAGE 

Stresses, composition of 676 

conjugate 676 

equilibrium of 668 

greatest, in girder 183 

in roof-trusses, determination of . .150 
mode of ascertaining, in a beam . .281 

parallel to a plane, composition . . 679 

principal 681 

principal, determination of ... . 682 

Stretching and tearing 232 

Strut 138 

Struts, short 313 

Suspension-rod of uniform strength . . 240 

Tensile limit, Beardslee 368 

Tension of timber 497 

Theory of elasticity 658 

Tie 138 

Timber beams, immediate modulus of 

elasticity 535 

beams, modulus of elasticity for 

use . 536 

beams, strength and deflection . .523 

beams, time tests 534 

columns 498 

columns loaded eccentrically . . . 506 

columns, seasoned 505 

compression 498 

effect of seasoning 548 

factor of safety 522 

floors 546 

framing-joints ........ 546 

general remarks 488 

list of experimenters 490 

longitudinal shearing in beams . . 537 
posts, rules for strength, Ely . . .521 

shearing 547 

strength as giv"en by Rankine . . .491 
strength as given by Rodman . . . 495 
strength, general remarks .... 489 

tension 497 

transverse strength 523 

Time of descent down a curve .... 96 

Time tests on timber beams 534 

Torsional strength of iron and steel . . 438 
Torsion and bending combined .... 690 



INDEX. 



729 



PAGE 

Translation and rotation combined . . 44 
Transverse deflection of shafts . . . .327 
Transverse strength of cast-iron .... 349 

steel 425 

timber 523 

wrought-iron 397 

Travelling-load 182 

Triangular frame 141 

Triangular truss, wind pressure .... 147 

Truss, Bollman's 209 

Fink's 207 

hammer-beam 166 

scissor-beam 169 

triangular, wind pressure .... 147 

Trusses, bridge 174 

methods for determining stresses . 140 

roof 138, 156 

roof, determination of stresses . .150 
Twisting and bending combined . . . 328 

Velocity 2, 75 

Wade, tests of cast-iron 341 

Watertown Arsenal, steel eye-bars . . .417 

tensile strength and elasticity of 
steel 412 

tensile strength and elasticity of 
wrought-iron 374 

tests of bridge columns 389 

tests of riveted joints 455 

Weight of materials for roofs . . . .151 

of snow 151 

Weyrauch's formula 245, 432 



PAGE 

White-pine beams . . 537 

columns 511 

Wind pressure 152 

triangular truss 147 

Wire, iron and steel 483 

rope 484 

Wdhler's experiments 427 

on compressive strength 244 

on tensile strength 237 

results 434 

Working-strength 235 

of beams 283 

Work, mechanical yy 

under oblique force 104 

unit of yy 

Wrought-iron beams, strength and deflec- 
tion 397 

Wrought-iron, characteristics 351 

columns, tests, full-size 383 

compressive strength 382 

eye-bars 376 

list of experimenters 352 

tensile and compressive strength . . 352 
tensile strength and elasticity, Hodg- 

kinson 353 

tensile tests at Watertown Arsenal . 374 

tests by Beardslee 367 

tests by Kirkaldy 356 

Transverse strength 397 

Yellow-pine beams, strength and elas- 

1 ticity 532 

columns, tables 503, 516 



